Limits, Continuity, Derivatives

I nformal Definition of Limit

If the values taken on by a function $ f(x) $ can be made arbitrarily close to L by taking input values x arbitrarily close to a, then L is called the limit of $ f(x) $ as x approaches a, written

\[ \lim_{x\to a}f(x)=L \]

EXAMPLE 45.1 $ _{x}(2x-3)=5 $ , since 2x-3 can be made arbitrarily close to 5 by taking values of x arbitrarily close to 4, as suggested by the following table:

x 3.5 3.9 3.99 3.999 4.5 4.1 4.01 4.001
2x - 3 4 4.8 4.98 4.998 6 5.2 5.02 5.002

Formal Definition of Limit

$ _{xa}f(x)=L $ means that given any $ >0 $ , a number $ >0 $ can be found so that if $ 0<|x-a|<$ , then $ |f(x)-L|<$ .

EXAMPLE 45.2 In the previous example, given any \(\varepsilon > 0\), take \(0 < |x - 4| < \varepsilon / 2\).

Then:

\[ \begin{aligned}2|x-4|&<\varepsilon\\|2x-8|&<\varepsilon\\|(2x-3)-5|&<\varepsilon\end{aligned} \]

Thus $ _{x}(2x-3)=5 $ .

Note that the statement $ _{xa}f(x)=L $ says nothing about what happens at a. Possibly $ f(a)=L $ ; however, possibly $ f(a) $ is undefined, or defined but unequal to L.

Properties of Limits

\[ \lim_{x\to a}c=c\qquad\lim_{x\to a}x=a \]

If $ {xa}f(x)=L $ and $ {xa}g(x)=M $ , then

\[ \lim_{x\to a}[f(x)+g(x)]=L+M\qquad\lim_{x\to a}[f(x)-g(x)]=L-M \]

\[ \lim_{x\to a}[f(x)g(x)]=LM\quad\lim_{x\to a}[f(x)]^{n}=L^{n} \]

\[ \lim_{x\to a}[f(x)/g(x)]=L/M\quad provided\quad M\neq0 \]

$ _{xa}= $ provided n is an odd integer, or n is an even integer and L is positive.

Finding Limits Algebraically

As a result of these properties, many limits can be found algebraically.

EXAMPLE 45.3 Find $ _{x }(3x + 7) $ .

\[ \lim_{x\to4}(3x+7)=\lim_{x\to4}3x+\lim_{x\to4}7=\lim_{x\to4}3\cdot\lim_{x\to4}x+\lim_{x\to4}7=3\cdot4+7=19 \]

EXAMPLE 45.4 Find $ _{x } $

\[ \lim_{x\to3}\frac{x^{2}-9}{x-3}=\lim_{x\to3}\frac{(x-3)(x+3)}{x-3}=\lim_{x\to3}(x+3)=\lim_{x\to3}x+\lim_{x\to3}3=3+3=6 \]

There are, however, many situations where a limit does not exist.

EXAMPLE 45.5 Find $ _{x}(-) $ .

Consider the following table:

x -0.5 -0.1 -0.01 -0.001 0.5 0.1 0.01 0.001
$ - $ 2 10 100 1000 -2 -10 -100 -1000

The values are not approaching a limit; the limit does not exist.

One-Sided Limits

  1. If the values taken on by a function $ f(x) $ can be made arbitrarily close to L by taking input values x arbitrarily close to (but greater than) a, then L is called the limit of $ f(x) $ as x approaches a from the right, written

\[ \lim_{x\to a^{+}}f(x)=L \]

  1. If the values taken on by a function $ f(x) $ can be made arbitrarily close to L by taking input values x arbitrarily close to (but less than) a, then L is called the limit of $ f(x) $ as x approaches a from the left, written

\[ \lim_{x\to a^{-}}f(x)=L \]

I nfinite Limits

If the values taken on by a function $ f(x) $ can be made arbitrarily large and positive by taking input values x arbitrarily close to a, then it is said that the limit of $ f(x) $ as x approaches a is (positive) infinite, written $ {x a} f(x) = $ . If the values taken on by a function $ f(x) $ can be made arbitrarily large and negative by taking input values x arbitrarily close to a, then it is said that the limit of $ f(x) $ as x approaches a is negative infinite, written $ {x a} f(x) = -$ .

EXAMPLE 45.6 \(\lim_{x\to3}\frac{1}{(x-3)^{2}}=\infty\), since \(\frac{1}{(x-3)^{2}}\) can be made arbitrarily large by taking values of \(x\) arbitrarily close to 3, as suggested by the following table:

x 2.5 2.9 2.99 2.999 3.5 3.1 3.01 3.001
$ $ 4 100 10,000 1,000,000 4 100 10,000 1,000,000

One-Sided Infinite Limits

  1. If the values taken on by a function \(f(x)\) can be made arbitrarily large and positive by taking input values \(x\) arbitrarily close to (but greater than) \(a\), it is said that the limit of \(f(x)\) as \(x\) approaches \(a\) from the right is (positive) infinite, written \(\lim_{x\to a^{+}}f(x)=\infty\). If the values taken on by a function \(f(x)\) can be made arbitrarily large and negative by taking input values \(x\) arbitrarily close to (but greater than) \(a\), it is said that the limit of \(f(x)\) as \(x\) approaches \(a\) from the right is negative infinite, written \(\lim_{x\to a^{-}}f(x)=-\infty\).

  2. If the values taken on by a function $ f(x) $ can be made arbitrarily large and positive by taking input values x arbitrarily close to (but less than) a, it is said that the limit of $ f(x) $ as x approaches a from the left is (positive) infinite, written $ {x a} f(x) = $ . If the values taken on by a function $ f(x) $ can be made arbitrarily large and negative by taking input values x arbitrarily close to (but less than) a, it is said that the limit of $ f(x) $ as x approaches a from the left is negative infinite, written $ {x a} f(x) = -$ .

EXAMPLE 45.7 Find $ {x^{+}}(-) $ and $ {x^{-}}(-) $ .

From the table in Example 45.5, it appears that $ {x^{+}}(-)=-$ and $ {x^{-}}(-)=$ .

Limits at Infinity

  1. If the values taken on by a function $ f(x) $ can be made arbitrarily close to L by taking input values x arbitrarily large and positive, then L is called the limit of $ f(x) $ as x approaches (positive) infinity, written

\[ \lim_{x\to\infty}f(x)=L \]

  1. If the values taken on by a function $ f(x) $ can be made arbitrarily close to L by taking input values x arbitrarily large and negative, then L is called the limit of $ f(x) $ as x approaches negative infinity, written

\[ \lim_{x\to-\infty}f(x)=L \]

Definition of Continuity

  1. A function $ f(x) $ is called continuous for a value c if $ _{x c} f(x) = f(c) $ . This is usually referred to as continuity at a point c, or simply continuity at c.

  2. A function $ f(x) $ is called continuous on an open interval $ (a, b) $ if it is continuous at every point on the interval.

  3. A function $ f(x) $ is called continuous on a closed interval $ [a, b] $ if it is continuous at every point on the interval $ (a, b) $ and also $ {x a^{+}} f(x) = f(a) $ and $ {x b^{-}} f(x) = f(b) $ .

If a function is not continuous for a value c, it is called discontinuous, and c is called a point of discontinuity.

EXAMPLE 45.8 It can be shown that every polynomial function is continuous at every point in R and that every rational function is continuous at every point in its domain. Thus if $ f(x) $ is a polynomial function, the limit $ {xc}f(x) $ can always be calculated as $ f(c) $ . If $ f(x)=p(x)/q(x) $ is a rational function, the limit $ {xc}f(x) $ can be calculated as $ f(c)=p(c)/q(c) $ for any value c as long as $ f(c) $ is defined, that is, if $ q(c) $ .

Definition of Derivative

Given a function $ f(x) $ , the derivative of f, written $ f’(x) $ , is a function defined by the formula

\[ f^{\prime}(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h} \]

provided that the limit exists. If the limit exists for a value a (also referred to as: at the point a), the function is called differentiable at a. The process of finding the derivative is called differentiation.

EXAMPLE 45.9 Find the derivative of $ f(x) = x^{2} $

\[ \begin{aligned}\lim_{h\to0}\frac{f(x+h)-f(x)}{h}&=\lim_{h\to0}\frac{(x+h)^{2}-x^{2}}{h}\\&=\lim_{h\to0}\frac{x^{2}+2xh+h^{2}-x^{2}}{h}\\&=\lim_{h\to0}\frac{2xh+h^{2}}{h}\\&=\lim_{h\to0}\frac{h(2x+h)}{h}\\&=\lim_{h\to0}(2x+h)\\&=2x\end{aligned} \]

Average and Instantaneous Rates of Change

In Chapter 9, the average rate of change of $ f(x) $ over an interval from x to $ x + h $ was defined as

\[ \frac{f(x+h)-f(x)}{h} \]

also referred to as the difference quotient.

The derivative of $ f(x) $ , $ f’(x) = _{h } $ , is also called the instantaneous rate of change of the function with respect to the variable x.

Tangent Line

The tangent line to the graph of a function $ f(x) $ at the point $ (a, f(a)) $ is the straight line through the point with slope m equal to the derivative of the function at the point a,

\[ m(a)=f^{\prime}(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h} \]

Average and Instantaneous Velocity

Given a function $ s(t) $ that represents the position of an object at time t, the average velocity of the object on the interval $ [a, b] $ is given by

\[ \frac{Change in position}{Change in time}=\frac{s(b)-s(a)}{b-a} \]

The instantaneous velocity of the object at time t is given by the derivative of $ s(t) $ :

\[ v(t)=s^{\prime}(t)=\lim_{h\to0}\frac{s(t+h)-s(t)}{h} \]

SOLVED PROBLEMS

45.1. Use the formal definition of limit to show (a) $ {x a} c = c $ ; (b) $ {x a} x = a $ .

  1. $ _{xa}c = c $ means that given any $ > 0 $ a number $ > 0 $ can be found so that if $ 0 < |x - a| < $ , then $ |c - c| < $ . However, regardless of x and $ $ , $ |c - c| = |0| = 0 < $ holds for any $ > 0 $ . This proves the required result.

  2. $ _{xa}x=a $ means that given any $ >0 $ , a number $ >0 $ can be found so that if $ 0<|x-a|<$ , then $ |x-a|<$ . Clearly, given $ >0 $ , choose $ =$ , then $ 0<|x-a|<$ will guarantee $ |x-a|<$ . This proves the required result.

45.2. Use the formal definition of limit to show that if $ {xa}f(x)=L $ and $ {xa}g(x)=M $ , then $ _{xa}[f(x)+g(x)]=L+M $ .

$ _{xa}[f(x)+g(x)]=L+M $ means that given any $ >0 $ , a number $ >0 $ can be found so that if $ 0<|x-a|<$ , then $ |(f(x)+g(x))-(L+M)|<$ .

Note that $ |(f(x) + g(x)) - (L + M)| = |(f(x) - L) + (g(x) - M)| |f(x) - L| + |g(x) - M| $ . This inequality follows from the triangle inequality (Chapter 7).

Hence, since \(\lim_{x\to a}f(x)=L\) and \(\lim_{x\to a}g(x)=M\), given \(\varepsilon>0\), choose \(\delta_{1}\) so that if \(0<|x-a|<\delta_{1}\), then \(|f(x)-L|<\varepsilon/2\), and choose \(\delta_{2}\) so that if \(0<|x-a|<\delta_{2}\) then \(|g(x)-M|<\varepsilon/2\). Therefore, choose \(\delta\) to be the smaller of \(\delta_{1}\) and \(\delta_{2}\). Then, if \(0<|x-a|<\delta\),

$ |(f(x) + g(x)) - (L + M)| |f(x) - L| + |g(x) - M| < / 2 + / 2 = $ . This proves the required result.

45.3. Find $ _{x}(2-3x) $ (a) by examining a table of values near 5; (b) by using the formal definition of limit; (c) by using the continuity of polynomial functions.

  1. Form a table of values near 5:
x 4.5 4.9 4.99 4.999 5.5 5.1 5.01 5.001
2 - 3x -11.5 -12.7 -12.97 -12.997 -14.5 -13.3 -13.03 -13.003

This suggests that $ _{x}(2-3x)=-13 $ .

  1. If $ 0 < |x - 5| < $ , then

\[ 0<|-3(x-5)|<|-3|\delta_{1} \]

\[ 0<\left|15-3x\right|<3\delta_{1} \]

\[ 0<\left|(2-3x)-(-13)\right|<3\delta_{1} \]

Therefore, given ε > 0, choose δ = δ₁ = ε/3. If 0 < |x - 5| < δ, then

\[ \left|(2-3x)-(-13)\right|<3(\varepsilon/3)=\varepsilon \]

Thus $ _{x}(2-3x)=-13 $ as suggested by the table.

  1. Since $ f(x) = 2 - 3x $ is a polynomial function, $ _{x } f(x) = f(5) = 2 - 3 = -13 $ .

45.4. Find (a) $ {x}(x{3}-3x{2}+2x+8) $ ; (b) $ {x} $ .

  1. Since $ f(x) = x^{3} - 3x^{2} + 2x + 8 $ is a polynomial function,

\[ \lim_{x\to-2}f(x)=f(-2)=(-2)^{3}-3(-2)^{2}+2(-2)+8=-16 \]

  1. Since $ f(x) = $ is a rational function defined at x = 4,

\[ \lim_{x\to4}f(x)=f(4)=\frac{4-1}{4^{2}+3}=\frac{3}{19} \]

45.5. Find the following limits algebraically:

\[ \begin{aligned}(a)\lim_{x\to5}\frac{x-5}{x^{2}-25};(b)\lim_{x\to16}\frac{\sqrt{x}-4}{x-16};(c)\lim_{x\to0}\frac{(3-x)^{2}-9}{x}\end{aligned} \]

\[ \begin{aligned}(a)\lim_{x\to5}\frac{x-5}{x^{2}-25}=\lim_{x\to5}\frac{x-5}{(x-5)(x+5)}=\lim_{x\to5}\frac{1}{x+5}=\frac{1}{10}\end{aligned} \]

\[ \begin{aligned}(b)\lim_{x\rightarrow16}\frac{\sqrt{x}-4}{x-16}&=\lim_{x\rightarrow16}\frac{\sqrt{x}-4}{\left(\sqrt{x}-4)(\sqrt{x}+4\right)}=\lim_{x\rightarrow16}\frac{1}{\sqrt{x}+4}=\frac{\lim\limits_{x\rightarrow16}1}{\sqrt{\lim\limits_{x\rightarrow16}x}+\lim\limits_{x\rightarrow16}4}=\frac{1}{\sqrt{16}+4}=\frac{1}{8}\end{aligned} \]

\[ \begin{aligned}(c)\lim_{x\rightarrow0}\frac{(3-x)^{2}-9}{x}&=\lim_{x\rightarrow0}\frac{9-6x+x^{2}-9}{x}=\lim_{x\rightarrow0}\frac{-6x+x^{2}}{x}=\lim_{x\rightarrow0}\frac{x(-6+x)}{x}=\lim_{x\rightarrow0}(-6+x)=-6\end{aligned} \]

45.6. (a) Give formal definitions of \(\lim_{x\to a^{+}}f(x)=L\) and \(\lim_{x\to a^{-}}f(x)=M\). (b) Show that if \(\lim_{x\to a^{+}}f(x)=L\) and \(\lim_{x\to a^{-}}f(x)=L\), then \(\lim_{x\to a}f(x)=L\).

  1. \(\lim_{x\to a^{-}}f(x)=L\) means that given any \(\varepsilon>0\), a number \(\delta>0\) can be found so that if \(0<x-a<\delta\) then

\(|f(x)-L|<\varepsilon\). \(\lim_{x\to a^{-}}f(x)=M\) means that given any \(\varepsilon>0\), a number \(\delta>0\) can be found so that if \(0<a-x<\delta\), then \(|f(x)-M|<\varepsilon\).

  1. If $ {xa^{+}}f(x)=L $ and $ {xa^{-}}f(x)=L $ , then given any $ >0 $ a number $ {1}>0 $ can be found so that if $ 0<x-a<{1} $ , then $ |f(x)-L|<$ and a number $ {2}>0 $ can be found so that if $ 0<a-x<{2} $ , then $ |f(x)-L|<$ . So given any $ >0 $ , choose $ $ to be the smaller of $ {1} $ and $ {2} $ . Then if $ 0<|x-a|<$ , both $ 0<x-a<{1} $ and $ 0<a-x<{2} $ will hold and $ |f(x)-L|<$ as required.
45.7. Let $ f(x) = { \[\begin{array}{ll} x - 3 & \text{if } x < 2 \\ 6x & \text{if } x \geq 2 \end{array}\]

. $ . Find (a) $ {x ^{+}} f(x) $ ; (b) $ {x ^{-}} f(x) $ ; (c) $ _{x } f(x) $ .

  1. $ _{x{+}}f(x)=_{x{+}}6x=12 $

\[ (b)\lim_{x\to2^{-}}f(x)=\lim_{x\to2^{-}}(x-3)=-1 \]

  1. Since $ {x{+}}f(x)_{x{-}}f(x),{x}f(x) $ does not exist.
45.8. Let $ f(x) = { \[\begin{aligned} & x^{2} - 3 & \text{if } x < 2 \\ & \frac{1}{2}x & \text{if } x \geq 2 \end{aligned}\]

. $ . Find (a) $ {x ^{+}} f(x) $ ; (b) $ {x ^{-}} f(x) $ ; (c) $ _{x } f(x) $ .

  1. $ _{x{+}}f(x)=_{x{+}}x=1 $

\[ \begin{aligned}(b)\lim_{x\to2^{-}}f(x)=\lim_{x\to2^{-}}(x^{2}-3)=2^{2}-3=1\end{aligned} \]

  1. Since $ {x{+}}f(x)=_{x{-}}f(x)=1,{x}f(x)=1 $

45.9. Find (a) \(\lim_{x\to4}\sqrt{x}\); (b) \(\lim_{x\to-4}\sqrt{x}\).

  1. $ _{x}===2 $

  2. Since $ $ is not a real number for any value of x near -4, $ _{x} $ does not exist.

45.10. Let $ f(x) = $ . Find (a) $ {x ^{+}} f(x) $ ; (b) $ {x ^{-}} f(x) $ ; (c) $ _{x } f(x) $ .

Consider the following table:

x 1.5 1.9 1.99 1.999 2.5 2.1 2.01 2.001
$ $ -2 -10 -100 -1000 2 10 100 1000

  1. From the table, it appears that $ {x^{+}}f(x) $ does not exist; however, it can be said that $ {x^{+}}f(x)=$ .

  2. From the table, it appears that $ {x^{-}}f(x) $ does not exist; however, it can be said that $ {x^{-}}f(x)=-$ .

  3. Since $ {x{+}}f(x)_{x{-}}f(x),{x}f(x) $ does not exist.

45.11. Let $ f(x)= $ . Find (a) $ {x^{+}}f(x) $ ; (b) $ {x^{-}}f(x) $ ; (c) $ _{x}f(x) $ .

Consider the following table:

x 1.5 1.9 1.99 1.999 2.5 2.1 2.01 2.001
$ $ 18 490 49,900 4,999,000 22 510 50,100 5,001,000

  1. From the table, it appears that $ {x^{+}}f(x) $ does not exist; however, it can be said that $ {x^{+}}f(x)=$ .

  2. From the table, it appears that $ {x^{-}}f(x) $ does not exist; however, it can be said that $ {x^{-}}f(x)=$ .

  3. Since $ {x{+}}f(x)=_{x{-}}f(x)=$ , it can be said that $ {x}f(x)=$ .

45.12. It can be shown that $ {x}=0 $ and $ {x-}=0 $ for any positive value of n. Use these facts to find (a) $ {x} $ ; (b) $ {x-} $ .

\[ \begin{aligned}(b)\lim_{x\to-\infty}\frac{5}{x^{2}+6}&=\frac{\lim\limits_{x\to-\infty}(5/x^{2})}{\lim\limits_{x\to-\infty}1+\lim\limits_{x\to-\infty}(6/x^{2})}=\frac{5\lim\limits_{x\to-\infty}(1/x^{2})}{\lim\limits_{x\to-\infty}1+6\lim\limits_{x\to-\infty}(1/x^{2})}=\frac{5\cdot0}{1+6\cdot0}=0\end{aligned} \]

45.13. Discuss how a function may fail to be continuous for a particular value and give examples.

A function $ f(x) $ is continuous for a value a if $ _{x a} f(x) = f(a) $ . However:

  1. $ f(a) $ may be undefined. For example, consider $ f(x) = $ at x = 0, or $ f(x) = $ at any negative value of x, or $ f(x) = $ at x = 2.

  2. The limit $ _{xa}f(x) $ may fail to exist. For example, consider $ f(x)= $ at x=0 or $ f(x)=

    \[\begin{cases}x^{2}\text{ if }x<2\\3x\text{ if }x\geq2\end{cases}\]

    $ at x=2.

  3. The limit $ {xa}f(x) $ may exist and $ f(a) $ may be defined, but $ {xa}f(x)f(a) $ . For example, consider

\[ f(x) = \begin{cases} x^2 & if x \neq 2 \\ 3 & if x = 2 \end{cases} \text{at } x = 2, \text{where} \lim_{x \to 2} f(x) = 4, \text{but } f(2) = 3. \]

45.14. Analyze the derivative of the function $ f(x) = |x| $ .

Find $ _{h} $

\[ \lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{\left|x+h\right|-\left|x\right|}{h} \]

If x > 0, then for sufficiently small $ h, x + h > 0 $ , hence

\[ \lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{\left|x+h\right|-\left|x\right|}{h}=\lim_{h\to0}\frac{x+h-x}{h}=\lim_{h\to0}\frac{h}{h}=\lim_{h\to0}1=1 \]

If x < 0, then for sufficiently small $ h, x + h < 0 $ , hence

\[ \lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{\left|x+h\right|-\left|x\right|}{h}=\lim_{h\to0}\frac{-(x+h)-(-x)}{h}=\lim_{h\to0}\frac{-h}{h}=\lim_{h\to0}(-1)=-1 \]

If x = 0, however.

\[ \lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{\left|x+h\right|-\left|x\right|}{h}=\lim_{h\to0}\frac{\left|h\right|}{h} \]

This limit does not exist, since

\[ \lim_{h\to0^{+}}\frac{|h|}{h}=\lim_{h\to0^{+}}\frac{h}{h}=\lim_{h\to0^{+}}1=1but \]

\[ \lim_{h\to0^{-}}\frac{|h|}{h}=\lim_{h\to0^{-}}\frac{-h}{h}=\lim_{h\to0^{-}}(-1)=-1 \]

Summarizing, if $ f(x) = |x| $ , then $ f’(x) = \[\begin{cases} 1 & \text{if } x > 0 \\ -1 & \text{if } x < 0 \\ \text{undefined if } x = 0 \end{cases}\]

$

45.15. Find the derivatives of the following functions:

  1. Find $ _{h} $ .

\[ \begin{aligned}\lim_{h\to0}\frac{f(x+h)-f(x)}{h}&=\frac{(x+h)^{3}-x^{3}}{h}\\&=\lim_{h\to0}\frac{x^{3}+3x^{2}h+3xh^{2}+h^{3}-x^{3}}{h}\\&=\lim_{h\to0}\frac{3x^{2}h+3xh^{2}+h^{3}}{h}\\&=\lim_{h\to0}\frac{h(3x^{2}+3xh+h^{2})}{h}\\&=\lim_{h\to0}(3x^{2}+3xh+h^{2})\\&=3x^{2}\end{aligned} \]

  1. Find $ _{h} $ .

\[ \begin{aligned}\lim_{h\to0}\frac{f(x+h)-f(x)}{h}&=\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt{x}}{h}\\&=\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt{x}}{h}\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}\\&=\lim_{h\to0}\frac{x+h-x}{h(\sqrt{x+h}+\sqrt{x})}\\&=\lim_{h\to0}\frac{h}{h(\sqrt{x+h}+\sqrt{x})}\\&=\lim_{h\to0}\frac{1}{\sqrt{x+h}+\sqrt{x}}\\&=\frac{1}{2\sqrt{x}}\end{aligned} \]

This is defined as long as x > 0.

  1. Find $ _{h} $ .

\[ \begin{aligned}\lim_{h\to0}\frac{f(x+h)-f(x)}{h}&=\lim_{h\to0}\frac{1/(x+h)-1/x}{h}\\&=\lim_{h\to0}\frac{x-(x+h)}{hx(x+h)}\\&=\lim_{h\to0}\frac{-h}{hx(x+h)}\\&=\lim_{h\to0}\frac{-1}{x(x+h)}\\&=\frac{-1}{x^{2}}\end{aligned} \]

45.16. Find the equation of the line tangent to the graph of

  1. Using the derivative found in the previous problem, we find that the slope of the tangent line at (2,8) is $ f’(2) = 3 ^{2} = 12 $ . Using the point-slope form of the equation of a line, we find that the equation of the line through (2,8) with slope 12 is

\[ \begin{aligned}y-8&=12(x-2)\\y-8&=12x-24\\y&=12x-16\end{aligned} \]

  1. Using the derivative found in the previous problem, we find that the slope of the tangent line at (4,2) is $ f’(4) = = $ . Using the point-slope form of the equation of a line, we find that the equation of the line through (4,2) with slope $ $ is

\[ \begin{aligned}y-2&=\frac{1}{4}(x-4)\\y-2&=\frac{1}{4}x-1\\y&=\frac{1}{4}x+1\end{aligned} \]

  1. Using the derivative found in the previous problem, we find that the slope of the tangent line at $ (-1,-1) $ is $ f’(-1) = = -1 $ . Using the point-slope form of the equation of a line, we find that the equation of the line through $ (-1,-1) $ with slope -1 is

\[ \begin{aligned}y-(-1)&=(-1)[x-(-1)]\\y+1&=-x-1\\y&=-x-2\end{aligned} \]

SUPPLEMENTARY PROBLEMS

45.17. Find the following limits algebraically:

  1. $ {x}(5x+1) $ ; (b) $ {x}(2x^{2}-8x+7) $ ; (c) $ _{x} $

Ans. (a) -14; (b) -1; (c) 1

45.18. Find the following limits algebraically:

\[ \lim_{x\to2}\frac{x^{2}-4}{2x-4};(b)\lim_{x\to100}\frac{\sqrt{x}-10}{x-100};(c)\lim_{x\to2}\frac{2x+4}{x^{2}-4}(d)\lim_{x\to0}\frac{(x-3)^{3}+27}{x} \]

Ans. (a) 2; (b) $ $ ; (c) does not exist; (d) 27

45.19. Use the formal definition of limit to prove:

\[ \mathrm{f}\lim_{x\to a}f(x)=L\mathrm{and}\lim_{x\to a}g(x)=M,\mathrm{then}\lim_{x\to a}[f(x)-g(x)]=L-M. \]

45.20. Let $ f(x)={ \[\begin{aligned}&x^{3}\text{ if }x<3\\ &3x\text{ if }x\geq3\end{aligned}\]

. $ . Find (a) $ {x^{+}}f(x) $ ; (b) $ {x^{-}}f(x) $ ; (c) $ _{x}f(x) $

Ans. (a) 9; (b) 27; (c) does not exist

45.21. Let $ f(x)={ \[\begin{aligned}&x^{3}-18&if x<3\\ &3x&if x\geq3\end{aligned}\]

. $ . Find (a) $ {x^{+}}f(x) $ ; (b) $ {x^{-}}f(x) $ ; (c) $ _{x}f(x) $

Ans. (a) 9; (b) 9; (c) 9

45.22. Find the following limits:

  1. $ {x} $ ; (b) $ {x} $

Ans. (a) does not exist; (b) 3

45.22. Let $ f(x) = $ . Find (a) $ {x ^{+}} f(x) $ ; (b) $ {x ^{-}} f(x) $ ; (c) $ _{x } f(x) $ .

Ans. (a) $ $ ; (b) $ -$ ; (c) does not exist

45.23. Let $ f(x)= $ . Find (a) $ {x^{+}}f(x) $ ; (b) $ {x^{-}}f(x) $ ; (c) $ _{x}f(x) $ .

Ans. (a) $ -$ ; (b) $ -$ ; (c) $ -$

45.24. Find the following limits:

\[ \begin{aligned}(a)\lim_{x\to\infty}\frac{100x}{5x^{2}-1};(b)\lim_{x\to-\infty}\frac{100x^{2}}{5x^{2}-1}\end{aligned} \]

Ans. (a) 0; (b) 20

45.25. Show that if $ f(x) = mx + b $ , where m and b are constants, then $ f’(x) = m $ .

45.26. Find the derivatives of the following functions:

\[ Ans.\quad(a)f^{\prime}(x)=4x^{3};(b)f^{\prime}(x)=\frac{1}{2\sqrt{x-4}},x>4;(c)f^{\prime}(x)=\frac{-2}{x^{3}} \]

45.27. Find the equation of the line tangent to the graph of

\[ \left(a\right)f(x)=x^{4},at\left(1,1\right);\left(b\right)f(x)=\sqrt{x-4},at\left(5,1\right);\left(c\right)f(x)=\frac{1}{x^{2}},at\left(\frac{1}{2},4\right) \]

\[ Ans.\quad(a)y=4x-3;(b)y=\frac{1}{2}x-\frac{3}{2};(c)y=-16x+12 \]

45.28. Show that if $ f(x) = x^{n} $ , where n is any positive integer, then $ f’(x) = nx^{n-1} $ .

(Hint: Use the binomial theorem.)

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