2.4 Connections F, F' And Trigap Calculus

If f is continuous on a, b, and if F is any antiderivative of f on a, b, then b a ∫ f tdt = Fb Fa −. Note: These two theorems may be presented in reverse order. Part II is sometimes called the Integral Evaluation Theorem. Don’t overlook the obvious! a a d f tdt dx ∫ = 0, because the definite. 4.4 Areas, Integrals and Antiderivatives Contemporary Calculus 3 To evaluate a definite integral ⌡ ⌠ a b f(t) dt, we can find any antiderivative F of f and evaluate F(b) – F(a). This result is a special case of Part 2 of the Fundamental Theorem of Calculus, and it will be used hundreds of times in the next several chapters. 4.3 Connecting f ' and f ' with the graph of f Calculus Using the results from the last few statements leads us to the following definition. Definition of Concavity Let y = f (x) be a differentiable function on an interval I.

  1. 2.4 Connections F F' And Trigap Calculus Worksheet
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  3. 2.4 Connections F F' And Trigap Calculus Solutions

So f prime of a is equal to 0. And if f prime of a is equal to 0 and if we're concave upwards in the interval around a, so if the second derivative is greater than 0, then it's pretty clear, you see here, that we are dealing with a minimum point at a. Name: Block: Date: AP Calculus 4.3c Even More Connecting f ′ and f ′ ′ with the Graph of f Example 1 More on moving from the graph of f ′ to draw a possible graph of f a) The graph of f ′ is shown below (Note: This is the graph of the derivative of f) a) Over what intervals is the graph of f increasing?

Learning Objectives

  • Define the derivative function of a given function.
  • Graph a derivative function from the graph of a given function.
  • State the connection between derivatives and continuity.
  • Describe three conditions for when a function does not have a derivative.
  • Explain the meaning of a higher-order derivative.

As we have seen, the derivative of a function at a given point gives us the rate of change or slope of the tangent line to the function at that point. If we differentiate a position function at a given time, we obtain the velocity at that time. It seems reasonable to conclude that knowing the derivative of the function at every point would produce valuable information about the behavior of the function. However, the process of finding the derivative at even a handful of values using the techniques of the preceding section would quickly become quite tedious. In this section we define the derivative function and learn a process for finding it.

Derivative Functions

The derivative function gives the derivative of a function at each point in the domain of the original function for which the derivative is defined. We can formally define a derivative function as follows.

Definition: Derivative Function

Let (f) be a function. The derivative function, denoted by (f'), is the function whose domain consists of those values of (x) such that the following limit exists:

[f'(x)=lim_{h→0}frac{f(x+h)−f(x)}{h}. label{derdef}]

A function (f(x)) is said to be differentiableat (a) if (f'(a)) exists. More generally, a function is said to be differentiableon (S) if it is differentiable at every point in an open set (S), and a differentiable function is one in which (f'(x)) exists on its domain.

In the next few examples we use Equation ref{derdef} to find the derivative of a function.

Example (PageIndex{1}): Finding the Derivative of a Square-Root Function

Find the derivative of (f(x)=sqrt{x}).

Solution

Start directly with the definition of the derivative function.

Substitute (f(x+h)=sqrt{x+h}) and (f(x)=sqrt{x}) into (f'(x)= displaystyle lim_{h→0}frac{f(x+h)−f(x)}{h}).

(f'(x)=displaystyle lim_{h→0}frac{sqrt{x+h}−sqrt{x}}{h})
(=displaystylelim_{h→0}frac{sqrt{x+h}−sqrt{x}}{h}⋅frac{sqrt{x+h}+sqrt{x}}{sqrt{x+h}+sqrt{x}})Multiply numerator and denominator by (sqrt{x+h}+sqrt{x}) without distributing in the denominator.
(=displaystylelim_{h→0}frac{h}{hleft(sqrt{x+h}+sqrt{x}right)})Multiply the numerators and simplify.
(=displaystylelim_{h→0}frac{1}{left(sqrt{x+h}+sqrt{x}right)})Cancel the (h).
(=dfrac{1}{2sqrt{x}})Evaluate the limit

Example (PageIndex{2}): Finding the Derivative of a Quadratic Function

Find the derivative of the function (f(x)=x^2−2x).

Solution

Follow the same procedure here, but without having to multiply by the conjugate.

2.4 Connections F, F

Substitute (f(x+h)=(x+h)^2−2(x+h)) and (f(x)=x^2−2x) into (f'(x)= displaystyle lim_{h→0}frac{f(x+h)−f(x)}{h}.)

(f'(x)=displaystylelim_{h→0}frac{((x+h)^2−2(x+h))−(x^2−2x)}{h})
(=displaystylelim_{h→0}frac{x^2+2xh+h^2−2x−2h−x^2+2x}{h})Expand ((x+h)^2−2(x+h)).
(=displaystylelim_{h→0}frac{2xh−2h+h^2}{h})Simplify
(=displaystylelim_{h→0}frac{h(2x−2+h)}{h})Factor out (h) from the numerator
(=displaystylelim_{h→0}(2x−2+h))Cancel the common factor of (h)
(=2x−2)Evaluate the limit

Exercise: (PageIndex{1})

Find the derivative of (f(x)=x^2).

Hint

Use Equation ref{derdef} and follow the example.

Answer

(f'(x)=2x)

We use a variety of different notations to express the derivative of a function. In Example we showed that if (f(x)=x^2−2x), then (f'(x)=2x−2). If we had expressed this function in the form (y=x^2−2x), we could have expressed the derivative as (y′=2x−2) or (dfrac{dy}{dx}=2x−2). We could have conveyed the same information by writing (dfrac{d}{dx}left(x^2−2xright)=2x−2). Thus, for the function (y=f(x)), each of the following notations represents the derivative of (f(x)):

(f'(x), quad dfrac{dy}{dx}, quad y′,quad dfrac{d}{dx}big(f(x)big)).

In place of (f'(a)) we may also use (dfrac{dy}{dx}Big _{x=a}). Use of the (dfrac{dy}{dx}) notation (called Leibniz notation) is quite common in engineering and physics. To understand this notation better, recall that the derivative of a function at a point is the limit of the slopes of secant lines as the secant lines approach the tangent line. The slopes of these secant lines are often expressed in the form (dfrac{Δy}{Δx}) where (Δy) is the difference in the (y) values corresponding to the difference in the (x) values, which are expressed as (Δx) (Figure (PageIndex{1})). Thus the derivative, which can be thought of as the instantaneous rate of change of (y) with respect to (x), is expressed as

(displaystyle frac{dy}{dx}= lim_{Δx→0}frac{Δy}{Δx}).

Graphing a Derivative

We have already discussed how to graph a function, so given the equation of a function or the equation of a derivative function, we could graph it. Given both, we would expect to see a correspondence between the graphs of these two functions, since (f'(x)) gives the rate of change of a function (f(x)) (or slope of the tangent line to (f(x))).

In Example (PageIndex{1}), we found that for (f(x)=sqrt{x}), (f'(x)=frac{1}{2sqrt{x}}). If we graph these functions on the same axes, as in Figure (PageIndex{2}), we can use the graphs to understand the relationship between these two functions. First, we notice that (f(x)) is increasing over its entire domain, which means that the slopes of its tangent lines at all points are positive. Consequently, we expect (f'(x)>0) for all values of x in its domain. Furthermore, as (x) increases, the slopes of the tangent lines to (f(x)) are decreasing and we expect to see a corresponding decrease in (f'(x)). We also observe that (f(0)) is undefined and that (displaystyle lim_{x→0^+}f'(x)=+∞), corresponding to a vertical tangent to (f(x)) at (0).

In Example (PageIndex{2}), we found that for (f(x)=x^2−2x,; f'(x)=2x−2). The graphs of these functions are shown in Figure (PageIndex{3}). Observe that (f(x)) is decreasing for (x<1). For these same values of (x), (f'(x)<0). For values of (x>1), (f(x)) is increasing and (f'(x)>0). Also, (f(x)) has a horizontal tangent at (x=1) and (f'(1)=0).

Example (PageIndex{3}): Sketching a Derivative Using a Function

Use the following graph of (f(x)) to sketch a graph of (f'(x)).

Solution

The solution is shown in the following graph. Observe that (f(x)) is increasing and (f'(x)>0) on ((–2,3)). Also, (f(x)) is decreasing and (f'(x)<0) on ((−∞,−2)) and on ((3,+∞)). Also note that (f(x)) has horizontal tangents at (–2) and (3), and (f'(−2)=0) and (f'(3)=0).

(PageIndex{2})

Sketch the graph of (f(x)=x^2−4). On what interval is the graph of (f'(x)) above the (x)-axis?

Hint

The graph of (f'(x)) is positive where (f(x)) is increasing.

Answer

((0,+∞))

Derivatives and Continuity

Now that we can graph a derivative, let’s examine the behavior of the graphs. First, we consider the relationship between differentiability and continuity. We will see that if a function is differentiable at a point, it must be continuous there; however, a function that is continuous at a point need not be differentiable at that point. In fact, a function may be continuous at a point and fail to be differentiable at the point for one of several reasons.

Differentiability Implies Continuity

Let (f(x)) be a function and (a) be in its domain. If (f(x)) is differentiable at (a), then (f) is continuous at (a).

Proof

If (f(x)) is differentiable at (a), then (f'(a)) exists and, if we let (h = x - a), we have ( x = a + h ), and as (h=x-ato 0), we can see that (xto a).

Then

[ f'(a) = lim_{hto 0}frac{f(a+h)-f(a)}{h}nonumber]

can be rewritten as

(f'(a)=displaystyle lim_{x→a}frac{f(x)−f(a)}{x−a}).

We want to show that (f(x)) is continuous at (a) by showing that (displaystyle lim_{x→a}f(x)=f(a).) Thus,

(begin{align*} displaystyle lim_{x→a}f(x) &=lim_{x→a};big(f(x)−f(a)+f(a)big)[4pt]
&=lim_{x→a}left(frac{f(x)−f(a)}{x−a}⋅(x−a)+f(a)right) & & text{Multiply and divide }(f(x)−f(a))text{ by }x−a.[4pt]
&=left(lim_{x→a}frac{f(x)−f(a)}{x−a}right)⋅left( lim_{x→a};(x−a)right)+lim_{x→a}f(a)[4pt]
&=f'(a)⋅0+f(a)[4pt]
&=f(a). end{align*})

Therefore, since (f(a)) is defined and (displaystyle lim_{x→a}f(x)=f(a)), we conclude that (f) is continuous at (a).

We have just proven that differentiability implies continuity, but now we consider whether continuity implies differentiability. To determine an answer to this question, we examine the function (f(x)= x ). This function is continuous everywhere; however, (f'(0)) is undefined. This observation leads us to believe that continuity does not imply differentiability. Let’s explore further. For (f(x)= x ),

(f'(0)=displaystyle lim_{x→0}frac{f(x)−f(0)}{x−0}= lim_{x→0}frac{ x − 0 }{x−0}= lim_{x→0}frac{ x }{x}).

This limit does not exist because

(displaystyle lim_{x→0^−}frac{ x }{x}=−1) and (displaystyle lim_{x→0^+}frac{ x }{x}=1).

See Figure (PageIndex{4}).

Let’s consider some additional situations in which a continuous function fails to be differentiable. Consider the function (f(x)=sqrt[3]{x}):

(f'(0)=displaystyle lim_{x→0}frac{sqrt[3]{x}−0}{x−0}=displaystyle lim_{x→0}frac{1}{sqrt[3]{x^2}}=+∞).

Thus (f'(0)) does not exist. A quick look at the graph of (f(x)=sqrt[3]{x}) clarifies the situation. The function has a vertical tangent line at (0) (Figure (PageIndex{5})).

The function (f(x)=begin{cases} xsinleft(frac{1}{x}right), & & text{ if } x≠00, & & text{ if } x=0end{cases}) also has a derivative that exhibits interesting behavior at (0).

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We see that

(f'(0)=displaystyle lim_{x→0}frac{xsinleft(1/xright)−0}{x−0}= lim_{x→0}sinleft(frac{1}{x}right)).

This limit does not exist, essentially because the slopes of the secant lines continuously change direction as they approach zero (Figure (PageIndex{6})).

In summary:

  1. We observe that if a function is not continuous, it cannot be differentiable, since every differentiable function must be continuous. However, if a function is continuous, it may still fail to be differentiable.
  2. We saw that (f(x)= x ) failed to be differentiable at (0) because the limit of the slopes of the tangent lines on the left and right were not the same. Visually, this resulted in a sharp corner on the graph of the function at (0.) From this we conclude that in order to be differentiable at a point, a function must be “smooth” at that point.
  3. As we saw in the example of (f(x)=sqrt[3]{x}), a function fails to be differentiable at a point where there is a vertical tangent line.
  4. As we saw with (f(x)=begin{cases}xsinleft(frac{1}{x}right), & & text{ if } x≠00, & &text{ if } x=0end{cases}) a function may fail to be differentiable at a point in more complicated ways as well.

Example (PageIndex{4}): A Piecewise Function that is Continuous and Differentiable

A toy company wants to design a track for a toy car that starts out along a parabolic curve and then converts to a straight line (Figure (PageIndex{7})). The function that describes the track is to have the form (f(x)=begin{cases}frac{1}{10}x^2+bx+c, & & text{ if }x<−10−frac{1}{4}x+frac{5}{2}, & & text{ if } x≥−10end{cases}) where (x) and (f(x)) are in inches. For the car to move smoothly along the track, the function (f(x)) must be both continuous and differentiable at (−10). Find values of (b) and (c) that make (f(x)) both continuous and differentiable.

Solution

For the function to be continuous at (x=−10), (displaystyle lim_{x→10^−}f(x)=f(−10)). Thus, since

(displaystyle lim_{x→−10^−}f(x)=frac{1}{10}(−10)^2−10b+c=10−10b+c)

and (f(−10)=5), we must have (10−10b+c=5). Equivalently, we have (c=10b−5).

For the function to be differentiable at (−10),

(f'(10)=displaystyle lim_{x→−10}frac{f(x)−f(−10)}{x+10})

must exist. Since (f(x)) is defined using different rules on the right and the left, we must evaluate this limit from the right and the left and then set them equal to each other:

(displaystyle begin{align*} lim_{x→−10^−}frac{f(x)−f(−10)}{x+10} &= lim_{x→−10^−}frac{frac{1}{10}x^2+bx+c−5}{x+10}[4pt]
&= lim_{x→−10^−}frac{frac{1}{10}x^2+bx+(10b−5)−5}{x+10} & & text{Substitute }c=10b−5.[4pt]
&= lim_{x→−10^−}frac{x^2−100+10bx+100b}{10(x+10)}[4pt]
&= lim_{x→−10^−}frac{(x+10)(x−10+10b)}{10(x+10)} & & text{Factor by grouping}[4pt]
&=b−2 end{align*}).

We also have

(displaystyle begin{align*} lim_{x→−10^+}frac{f(x)−f(−10)}{x+10} &= lim_{x→−10^+}frac{−frac{1}{4}x+frac{5}{2}−5}{x+10}[4pt]
&= lim_{x→−10^+}frac{−(x+10)}{4(x+10)}[4pt]
&=−frac{1}{4} end{align*}).

This gives us (b−2=−frac{1}{4}). Thus (b=frac{7}{4}) and (c=10(frac{7}{4})−5=frac{25}{2}).

Exercise (PageIndex{3})

Find values of a and b that make (f(x)=begin{cases}ax+b, & & text{ if } x<3x^2, & & text{ if } x≥3end{cases}) both continuous and differentiable at (3).

Hint

Use Example (PageIndex{4}) as a guide.

Answer

(a=6) and (b=−9)

And

Higher-Order Derivatives

The derivative of a function is itself a function, so we can find the derivative of a derivative. For example, the derivative of a position function is the rate of change of position, or velocity. The derivative of velocity is the rate of change of velocity, which is acceleration. The new function obtained by differentiating the derivative is called the second derivative. Furthermore, we can continue to take derivatives to obtain the third derivative, fourth derivative, and so on. Collectively, these are referred to as higher-order derivatives. The notation for the higher-order derivatives of (y=f(x)) can be expressed in any of the following forms:

(f'(x),; f''(x),; f^{(4)}(x),; …; ,; f^{(n)}(x))

(y'(x),; y''(x),; y^{(4)}(x),; …; ,; y^{(n)}(x))

(dfrac{d^2y}{dx^2},;dfrac{d^3y}{dy^3},;dfrac{d^4y}{dy^4},;…;,;dfrac{d^ny}{dy^n}.)

It is interesting to note that the notation for (dfrac{d^2y}{dx^2}) may be viewed as an attempt to express (dfrac{d}{dx}left(dfrac{dy}{dx}right)) more compactly.

Analogously, (dfrac{d}{dx}left(dfrac{d}{dx}left(dfrac{dy}{dx}right)right)=dfrac{d}{dx}left(dfrac{d^2y}{dx^2}right)=dfrac{d^3y}{dx^3}).

Example (PageIndex{5}): Finding a Second Derivative

For (f(x)=2x^2−3x+1), find (f'(x)).

Solution

First find (f'(x)).

Substitute (f(x)=2x^2−3x+1) and (f(x+h)=2(x+h)^2−3(x+h)+1) into (f'(x)=displaystyle lim_{h→0}dfrac{f(x+h)−f(x)}{h}.)

(f'(x)=displaystyle lim_{h→0}frac{(2(x+h)^2−3(x+h)+1)−(2x^2−3x+1)}{h})
(=displaystyle lim_{h→0}frac{4xh+h^2−3h}{h})Simplify the numerator.
(=displaystyle lim_{h→0}(4x+h−3))Factor out the (h) in the numerator and cancel with the (h) in the denominator.
(=4x−3)Take the limit.

Next, find (f'(x)) by taking the derivative of (f'(x)=4x−3.)

(f'(x)=displaystyle lim_{h→0}frac{f'(x+h)−f'(x)}{h})Use (f'(x)=displaystyle lim_{h→0}frac{f(x+h)−f(x)}{h}) with (f ′(x)) in place of (f(x).)
(=displaystyle lim_{h→0}frac{(4(x+h)−3)−(4x−3)}{h})Substitute (f'(x+h)=4(x+h)−3) and (f'(x)=4x−3.)
(=displaystyle lim_{h→0}4)Simplify.
(=4)Take the limit.

(PageIndex{4})

Find (f'(x)) for (f(x)=x^2).

Hint

We found (f'(x)=2x) in a previous checkpoint. Use Equation ref{derdef} to find the derivative of (f'(x))

Answer

(f'(x)=2)

Example (PageIndex{6}): Finding Acceleration

The position of a particle along a coordinate axis at time (t) (in seconds) is given by (s(t)=3t^2−4t+1) (in meters). Find the function that describes its acceleration at time (t).

Solution

Since (v(t)=s′(t)) and (a(t)=v′(t)=s'(t)), we begin by finding the derivative of (s(t)):

(displaystyle begin{align*} s′(t) &= lim_{h→0}frac{s(t+h)−s(t)}{h}[4pt]
&=lim_{h→0}frac{3(t+h)^2−4(t+h)+1−(3t^2−4t+1)}{h}[4pt]
&=6t−4. end{align*})

Next,

(displaystyle begin{align*} s'(t)&= lim_{h→0}frac{s′(t+h)−s′(t)}{h}[4pt]
&=lim_{h→0}frac{6(t+h)−4−(6t−4)}{h}[4pt]
&=6. end{align*})

Thus, (a=6 ;text{m/s}^2).

(PageIndex{5})

For (s(t)=t^3), find (a(t).)

Hint

Use Example (PageIndex{6}) as a guide.

Answer

(a(t)=6t)

Key Concepts

  • The derivative of a function (f(x)) is the function whose value at (x) is (f'(x)).
  • The graph of a derivative of a function (f(x)) is related to the graph of (f(x)). Where (f(x)) has a tangent line with positive slope, (f'(x)>0). Where (f(x)) has a tangent line with negative slope, (f'(x)<0). Where (f(x)) has a horizontal tangent line, (f'(x)=0.)
  • If a function is differentiable at a point, then it is continuous at that point. A function is not differentiable at a point if it is not continuous at the point, if it has a vertical tangent line at the point, or if the graph has a sharp corner or cusp.
  • Higher-order derivatives are derivatives of derivatives, from the second derivative to the (n^{text{th}}) derivative.

Key Equations

Calculus
  • The derivative function

(f'(x)=displaystyle lim_{h→0}frac{f(x+h)−f(x)}{h})

Glossary

derivative function
gives the derivative of a function at each point in the domain of the original function for which the derivative is defined
differentiable at (a)
a function for which (f'(a)) exists is differentiable at (a)
differentiable on (S)
a function for which (f'(x)) exists for each (x) in the open set (S) is differentiable on (S)
differentiable function
a function for which (f'(x)) exists is a differentiable function
higher-order derivative
a derivative of a derivative, from the second derivative to the (n^{text{th}}) derivative, is called a higher-order derivative

Contributors and Attributions

  • Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.

  • Paul Seeburger (Monroe Community College) added explanation of the alternative definition of the derivative used in the proof of that differentiability implies continuity.

2.4 Connections F F' And Trigap Calculus Worksheet

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Section 2-4 : Limit Properties

The time has almost come for us to actually compute some limits. However, before we do that we will need some properties of limits that will make our life somewhat easier. So, let’s take a look at those first. The proof of some of these properties can be found in the Proof of Various Limit Properties section of the Extras chapter.

Properties

First, we will assume that (mathop {lim }limits_{x to a} fleft( x right)) and (mathop {lim }limits_{x to a} gleft( x right)) exist and that (c) is any constant. Then,

  1. (mathop {lim }limits_{x to a} left[ {cfleft( x right)} right] = cmathop {lim }limits_{x to a} fleft( x right))

    In other words, we can “factor” a multiplicative constant out of a limit.

  2. (mathop {lim }limits_{x to a} left[ {fleft( x right) pm gleft( x right)} right] = mathop {lim }limits_{x to a} fleft( x right) pm mathop {lim }limits_{x to a} gleft( x right))

    So, to take the limit of a sum or difference all we need to do is take the limit of the individual parts and then put them back together with the appropriate sign. This is also not limited to two functions. This fact will work no matter how many functions we’ve got separated by “+” or “-”.

  3. (mathop {lim }limits_{x to a} left[ {fleft( x right)gleft( x right)} right] = mathop {lim }limits_{x to a} fleft( x right),mathop {lim }limits_{x to a} gleft( x right))

    We take the limits of products in the same way that we can take the limit of sums or differences. Just take the limit of the pieces and then put them back together. Also, as with sums or differences, this fact is not limited to just two functions.

  4. (displaystyle mathop {lim }limits_{x to a} left[ {frac{{fleft( x right)}}{{gleft( x right)}}} right] = frac{{mathop {lim }limits_{x to a} fleft( x right)}}{{mathop {lim }limits_{x to a} gleft( x right)}}{rm{,}},{rm{provided }},mathop {lim }limits_{x to a} gleft( x right) ne 0)

    As noted in the statement we only need to worry about the limit in the denominator being zero when we do the limit of a quotient. If it were zero we would end up with a division by zero error and we need to avoid that.

  5. (mathop {lim }limits_{x to a} {left[ {fleft( x right)} right]^n} = {left[ {mathop {lim }limits_{x to a} fleft( x right)} right]^n},{mbox{where }}n{mbox{ is any real number}})

    In this property (n) can be any real number (positive, negative, integer, fraction, irrational, zero, etc.). In the case that (n) is an integer this rule can be thought of as an extended case of 3.

    For example, consider the case of (n = )2.

    [begin{align*}mathop {lim }limits_{x to a} {left[ {fleft( x right)} right]^2} & = mathop {lim }limits_{x to a} left[ {fleft( x right)fleft( x right)} right] & = mathop {lim }limits_{x to a} fleft( x right)mathop {lim }limits_{x to a} fleft( x right)hspace{0.5in}{mbox{using property 3}} & = {left[ {mathop {lim }limits_{x to a} fleft( x right)} right]^2}end{align*}]

    The same can be done for any integer (n).

  6. (mathop {lim }limits_{x to a} left[ {sqrt[n]{{fleft( x right)}}} right] = sqrt[n]{{mathop {lim }limits_{x to a} fleft( x right)}})

    This is just a special case of the previous example.

    [begin{align*}mathop {lim }limits_{x to a} left[ {sqrt[n]{{fleft( x right)}}} right] & = mathop {lim }limits_{x to a} {left[ {fleft( x right)} right]^{frac{1}{n}}} & = {left[ {mathop {lim }limits_{x to a} fleft( x right)} right]^{frac{1}{n}}} & = sqrt[n]{{mathop {lim }limits_{x to a} fleft( x right)}}end{align*}]
  7. (mathop {lim }limits_{x to a} c = c,c{mbox{ is any real number}})

    In other words, the limit of a constant is just the constant. You should be able to convince yourself of this by drawing the graph of (fleft( x right) = c).

  8. (mathop {lim }limits_{x to a} x = a)

    As with the last one you should be able to convince yourself of this by drawing the graph of (fleft( x right) = x).

  9. (mathop {lim }limits_{x to a} {x^n} = {a^n})

    This is really just a special case of property 5 using (fleft( x right) = x).

Note that all these properties also hold for the two one-sided limits as well we just didn’t write them down with one sided limits to save on space.

Let’s compute a limit or two using these properties. The next couple of examples will lead us to some truly useful facts about limits that we will use on a continual basis.

Example 1 Compute the value of the following limit. [mathop {lim }limits_{x to - 2} left( {3{x^2} + 5x - 9} right)] Show Solution

This first time through we will use only the properties above to compute the limit.

First, we will use property 2 to break up the limit into three separate limits. We will then use property 1 to bring the constants out of the first two limits. Doing this gives us,

[begin{align*}mathop {lim }limits_{x to - 2} left( {3{x^2} + 5x - 9} right) & = mathop {lim }limits_{x to - 2} 3{x^2} + mathop {lim }limits_{x to - 2} 5x - mathop {lim }limits_{x to - 2} 9 & = 3mathop {lim }limits_{x to - 2} {x^2} + mathop {5lim }limits_{x to - 2} x - mathop {lim }limits_{x to - 2} 9end{align*}]

We can now use properties 7 through 9 to actually compute the limit.

[begin{align*}mathop {lim }limits_{x to - 2} left( {3{x^2} + 5x - 9} right) & = 3mathop {lim }limits_{x to - 2} {x^2} + mathop {5lim }limits_{x to - 2} x - mathop {lim }limits_{x to - 2} 9 & = 3{left( { - 2} right)^2} + 5left( { - 2} right) - 9 & = - 7end{align*}]

Now, let’s notice that if we had defined

[pleft( x right) = 3{x^2} + 5x - 9]

then the proceeding example would have been,

[begin{align*}mathop {lim }limits_{x to - 2} pleft( x right) & = mathop {lim }limits_{x to - 2} left( {3{x^2} + 5x - 9} right) & = 3{left( { - 2} right)^2} + 5left( { - 2} right) - 9 & = - 7 & = pleft( { - 2} right)end{align*}]

In other words, in this case we see that the limit is the same value that we’d get by just evaluating the function at the point in question. This seems to violate one of the main concepts about limits that we’ve seen to this point.

In the previous two sections we made a big deal about the fact that limits do not care about what is happening at the point in question. They only care about what is happening around the point. So how does the previous example fit into this since it appears to violate this main idea about limits?

Despite appearances the limit still doesn’t care about what the function is doing at (x = - 2). In this case the function that we’ve got is simply “nice enough” so that what is happening around the point is exactly the same as what is happening at the point. Eventually we will formalize up just what is meant by “nice enough”. At this point let’s not worry too much about what “nice enough” is. Let’s just take advantage of the fact that some functions will be “nice enough”, whatever that means.

The function in the last example was a polynomial. It turns out that all polynomials are “nice enough” so that what is happening around the point is exactly the same as what is happening at the point. This leads to the following fact.

Fact

If (p(x)) is a polynomial then,

[mathop {lim }limits_{x to a} pleft( x right) = pleft( a right)]

By the end of this section we will generalize this out considerably to most of the functions that we’ll be seeing throughout this course.

Let’s take a look at another example.

Example 2 Evaluate the following limit. [mathop {lim }limits_{z to 1} frac{{6 - 3z + 10{z^2}}}{{ - 2{z^4} + 7{z^3} + 1}}] Show Solution

First notice that we can use property 4 to write the limit as,

[mathop {lim }limits_{z to 1} frac{{6 - 3z + 10{z^2}}}{{ - 2{z^4} + 7{z^3} + 1}} = frac{{mathop {lim }limits_{z to 1} 6 - 3z + 10{z^2}}}{{mathop {lim }limits_{z to 1} - 2{z^4} + 7{z^3} + 1}}]

Well, actually we should be a little careful. We can do that provided the limit of the denominator isn’t zero. As we will see however, it isn’t in this case so we’re okay.

Now, both the numerator and denominator are polynomials so we can use the fact above to compute the limits of the numerator and the denominator and hence the limit itself.

[begin{align*}mathop {lim }limits_{z to 1} frac{{6 - 3z + 10{z^2}}}{{ - 2{z^4} + 7{z^3} + 1}} & = frac{{6 - 3left( 1 right) + 10{{left( 1 right)}^2}}}{{ - 2{{left( 1 right)}^4} + 7{{left( 1 right)}^3} + 1}} & = frac{{13}}{6}end{align*}]

2.4 Connections F F' And Trigap Calculus Pdf

Notice that the limit of the denominator wasn’t zero and so our use of property 4 was legitimate.

In the previous example, as with polynomials, all we really did was evaluate the function at the point in question. So, it appears that there is a fairly large class of functions for which this can be done. Let’s generalize the fact from above a little.

Fact

Calculus

Provided (f(x)) is “nice enough” we have,

[mathop {lim }limits_{x to a} fleft( x right) = fleft( a right)hspace{0.5in}mathop {lim }limits_{x to {a^ - }} fleft( x right) = fleft( a right)hspace{0.5in}mathop {lim }limits_{x to {a^ + }} fleft( x right) = fleft( a right)]

Again, we will formalize up just what we mean by “nice enough” eventually. At this point all we want to do is worry about which functions are “nice enough”. Some functions are “nice enough” for all (x) while others will only be “nice enough” for certain values of (x). It will all depend on the function.

As noted in the statement, this fact also holds for the two one-sided limits as well as the normal limit.

Here is a list of some of the more common functions that are “nice enough”.

  • Polynomials are nice enough for all (x)’s.
  • If (displaystyle fleft( x right) = frac{{pleft( x right)}}{{qleft( x right)}}) then (f(x)) will be nice enough provided both (p(x)) and (q(x)) are nice enough and if we don’t get division by zero at the point we’re evaluating at.
  • (cos left( x right),sin left( x right)) are nice enough for all (x)’s
  • (sec left( x right),tan left( x right)) are nice enough provided (x ne ldots , - frac{{5pi }}{2}, - frac{{3pi }}{2},frac{pi }{2},frac{{3pi }}{2},frac{{5pi }}{2}, ldots ) In other words secant and tangent are nice enough everywhere cosine isn’t zero. To see why recall that these are both really rational functions and that cosine is in the denominator of both then go back up and look at the second bullet above.
  • (csc left( x right),cot left( x right)) are nice enough provided (x ne ldots , - 2pi , - pi ,0,pi ,2pi , ldots ) In other words cosecant and cotangent are nice enough everywhere sine isn’t zero.
  • (sqrt[n]{x}) is nice enough for all (x) if (n) is odd.
  • (sqrt[n]{x}) is nice enough for (x ge 0) if (n) is even. Here we require (x ge 0) to avoid having to deal with complex values.
  • ({a^x},{{bf{e}}^x}) are nice enough for all (x)’s.
  • ({log _b}x,ln x) are nice enough for (x > 0). Remember we can only plug positive numbers into logarithms and not zero or negative numbers.
  • Any sum, difference or product of the above functions will also be nice enough. Quotients will be nice enough provided we don’t get division by zero upon evaluating the limit.

The last bullet is important. This means that for any combination of these functions all we need to do is evaluate the function at the point in question, making sure that none of the restrictions are violated. This means that we can now do a large number of limits.

Example 3

2.4 Connections F F' And Trigap Calculus Solutions

Evaluate the following limit. [mathop {lim }limits_{x to 3} left( { - sqrt[5]{x} + frac{{{{bf{e}}^x}}}{{1 + ln left( x right)}} + sin left( x right)cos left( x right)} right)] Show Solution

This is a combination of several of the functions listed above and none of the restrictions are violated so all we need to do is plug in (x = 3) into the function to get the limit.

[begin{align*}mathop {lim }limits_{x to 3} left( { - sqrt[5]{x} + frac{{{{bf{e}}^x}}}{{1 + ln left( x right)}} + sin left( x right)cos left( x right)} right) & = - sqrt[5]{3} + frac{{{{bf{e}}^3}}}{{1 + ln left( 3 right)}} + sin left( 3 right)cos left( 3 right) & = {rm{8}}{rm{.185427271}}end{align*}]

Not a very pretty answer, but we can now do the limit.