Topic 7 Inverse Functions

7.1 Inverse Functions - Definition and Properties

Exercise 7.1 If f(x)=x5+x3+x, find f−1(3) and (f−1)′(3).

Solution.

Since f(f−1(x))=x, by change rule, we have (f−1)′(x)=1f′(f−1(x))

To find f′(f−1(3)), we first find f′: f′(x)=(x5+x3+x)′=5x4+3x2+1 which is positive over its domain. So f is a strictly increasing function and hence one-to-one.

By the definition of inverse function, f−1(3) is the solution to the equation f(x)=3. Note that f(1)=3, so f−1(3)=1.

Put the information together, we find that (f−1)′(3)=1f′(f−1(3))=1f′(1)=15+3+1=19.

Exercise 7.2 Find a formula for the inverse of the function f(x)=4x−12x+3.

Solution.

Let y=f−1(x). By the definition of f−1(x), we have x=f(f−1(x))=f(y)=4y−12y+3.

Solve for y from this equation, we get y=f−1(x)=−3x+12x−4.

Exercise 7.3 Find a formula for the inverse of the function f(x)=1+√2+3x.

Solution.

To get the inverse function, solve for y from the following equation x=1+√2+3y, we get f−1(x)=y=13(x−1)2−23.

Exercise 7.4 Find an explicit formula for f−1 where f(x)=x4+1,x≥0.

Solution.

Solve for y from the following equation x=y4+1, we get f−1(x)=y=4√x−1.

Remark: Since the domain of f is x≥0, the range of f−1 is y≥0. That’s why we take y=4√x−1.

Exercise 7.5 Find (f−1)′(a), where f(x)=2x3+3x2+7x+4,a=4.

Solution.

Solve f(x)=4, we get f−1(4)=x=0. As the derivative of f is f′(x)=6x2+3x+7, apply the formula (f−1)′(x)=1f′(f−1(x)), we get (f−1)′(4)=1f′(f−1(4))=1f′(0)=17.

Exercise 7.6 Find (f−1)′(a), where f(x)=x3+3sinx+2cosx,a=2.

Solution.

Solve f(x)=2, we get f−1(2)=x=0. As the derivative of f is f′(x)=3x2+3cosx−2sinx, apply the formula (f−1)′(x)=1f′(f−1(x)), we get (f−1)′(2)=1f′(f−1(2))=1f′(0)=13.

Exercise 7.7 If g is an increasing function such that g(2)=8 and g′(2)=5, calculate (g−1)′(8).

Solution.

Since g(2)=8, g−1(8)=2. By the formula (f−1)′(x)=1f′(f−1(x)), we get (g−1)′(8)=1g′(g−1(8))=1g′(2)=15.

Exercise 7.8 If f(x)=∫x3√1+t3dt, find (f−1)′(0).

Solution.

By Fundamental Theorem of Calculus, we know that f′(x)=√1+x3≥0 over its domain x≥−1. Then f(x) is increasing over [−1,∞]. By Fundamental Theorem of Calculus, f(3)=∫33√1+t3dt=0. So f−1(0)=3.

Therefore, (f−1)′(0)=1f′(f−1(0))=1f′(3)=1√28=√714.

7.2 The Natural Logarithmic Function

Exercise 7.9 Find the limit limx→∞[ln(2+x)−ln(1+x)].

Solution.

limx→∞[ln(2+x)−ln(1+x)]=limx→∞ln(2+x1+x)=ln(limx→∞2+x1+x)=ln1=0

Remark: In the second last step, we used the trick that 2+x1+x=2x+11x+1.

Exercise 7.10 Differentiate the function f(x)=√xlnx.

Solution.

Apply the chain rule, we get f′(x)=(√xlnx)′=lnx2√x+√xx=√x(lnx+2)2x.

Exercise 7.11 Differentiate the function f(x)=ln(sin2x).

Solution.

Apply the chain rule twice, we get f′(x)=1sin2x(sin2x)′=1sin2x(2sinx)(sinx)′=2tanx.

Exercise 7.12 Differentiate the function y=1lnx.

Solution.

Apply the chain rule, we get f′(x)=−1(lnx)2(lnx)′=−1x(lnx)2.

Exercise 7.13 Differentiate the function y=(lntanx)2.

Solution.

Apply the chain rule twice, we get f′(x)=2(lntanx)(lntanx)′=2(lntanx)tanx(tanx)′=2sec2x(lntanx)tanx.

Exercise 7.14 Find f′′(e), where f(x)=lnxx.

Solution.

Apply the quotient rule and simplify, we get f′(x)=1−lnxx2

Apply the quotient rule again and simplify, we get f″

Therefore, f''(e)=\dfrac{2\ln e-3}{e^3}=\dfrac{2\cdot 1-3}{e^3}=-\dfrac1{e^3}.

Exercise 7.15 Use logarithmic differentiation to find the derivative of the function y=\sqrt{\frac{x-1}{x^{4}+1}}.

Solution.

Apply \ln to both sides, we get \ln y=\ln\left(\sqrt{\frac{x-1}{x^{4}+1}}\right)=\dfrac12\ln(x-1)-\dfrac12\ln(x^4+1).

Differentiate both sides, we get \dfrac1y\cdot y'=\frac1{2(x-1)}-\frac{4x^3}{2(x^4+1)}.

Multiply both sides by y=\sqrt{\frac{x-1}{x^{4}+1}}, we get y'= \left(\frac1{2(x-1)}-\frac{4x^3}{2(x^4+1)}\right)\sqrt{\frac{x-1}{x^{4}+1}}

Exercise 7.16 Evaluate the integral \int_{2}^{4} \frac{3}{x} \mathrm{d} x.

Solution.

\begin{aligned} &\int_{2}^{4} \frac{3}{x} \mathrm{d} x\\ =& 3\int_{2}^{4} \frac{1}{x} \mathrm{d} x\\ =& 3(\ln|x|)|_2^4\\ =& 3(\ln4-\ln2)=3\ln2. \end{aligned}

Exercise 7.17 Evaluate the integral \int_{1}^{2} \frac{\mathrm{d} t}{8-3t}.

Solution.

We find the anti-derivative first: \begin{aligned} &\int \frac{\mathrm{d} t}{8-3t}\\ \overset{u=8-3t}{=}& -\frac13\int \frac{\mathrm{d} u}{u}\\ =& -\frac13(\ln|u|)+C\\ =&-\frac13(\ln|8-3t|)+C. \end{aligned}

By Fundamental Theorem of Calculus, we get \int_{1}^{2} \frac{\mathrm{d} t}{8-3t}=-\frac13(\ln|8-3\cdot 2|-\ln|8-3\cdot 1|)=\frac{\ln5-\ln2}{3}.

Exercise 7.18 Evaluate the integral \int \frac{\cos x}{2+\sin x} \mathrm{d} x.

Solution.

Let u=2+\sin x. We find that \mathrm{d} u=\cos x\mathrm{d}x. Then \int \frac{\cos x}{2+\sin x} \mathrm{d} x=\int\frac{1}{u}\mathrm{d}u=\ln|u|+C=\ln(2+\sin x)+C.

Remark: In the last step, we remove the absolute value sign because 2+\sin x\ge1.

7.3 The Natural Exponential Function

Exercise 7.19 Find the domain of the function f(x)=\sqrt{3-e^{2 x}}.

Solution.

The domain of the function is given by the inequality 3-e^{2x}\geq 0, which is equivalent to e^{2x}\geq 3.

As the natural logarithmic function is strict increasing, take \ln of both sides, we get \begin{aligned} 2x\geq& \ln 3\\ x\ge & \frac12\ln 3. \end{aligned}

Exercise 7.20 Find the limit \lim_{x \to \infty}\left(e^{-2 x} \cos x\right).

Solution.

Since -1\le \cos x\le 1 and e^{-2x}>0, we have the following inequality -e^{-2x}\le e^{-2x}\cos x \le e^{-2x}.

By Squeeze theorem and the fact that \lim_{x\to \infty}e^{-2 x}=0, we know that \lim_{x \to \infty}\left(e^{-2 x} \cos x\right)=0.

Exercise 7.21 Differentiate the function f(t)=\sin \left(e^{t}\right)+e^{\sin t}.

Solution.

Apply chain rule to both terms: \begin{aligned} f'(x)&=\left(\sin(e^{t})\right)'+\left(e^{\sin t}\right)'\\ &=\cos(e^t)e^t+e^{\sin t}(\cos t)\\ &=e^t\cos(e^t)+e^{\sin t}\cos t . \end{aligned}

Exercise 7.22 Differentiate the function y=\cos \left(\frac{1-e^{2 x}}{1+e^{2 x}}\right).

Solution.

Note that \frac{1-e^{2 x}}{1+e^{2 x}}=\frac{2}{1+e^{2 x}}-1.

Apply chain rule three times: \begin{aligned} y'&=-\sin\left(\frac{1-e^{2 x}}{1+e^{2 x}}\right)\cdot \left(\frac{2}{1+e^{2 x}}-1\right)'\\ &=-\sin\left(\frac{1-e^{2 x}}{1+e^{2 x}}\right)\cdot \left(-\frac{2}{(1+e^{2 x})^2}\right)\cdot\left(1+e^{2x}\right)'\\ &=\sin\left(\frac{1-e^{2 x}}{1+e^{2 x}}\right)\cdot \left(\frac{2}{(1+e^{2 x})^2}\right)\cdot 2e^{2x}\\ &=\frac{4e^{2x}}{(1+e^{2 x})^2}\sin\left(\frac{1-e^{2 x}}{1+e^{2 x}}\right). \end{aligned}

Exercise 7.23 Find an equation of the tangent line to the curve x e^{y}+y e^{x}=1 at the point (0,1).

Solution.

The slope of the tangent line is given by y'(0). So we first need to find y'(x).

Differentiate both side with respect to x, we get e^y+xe^{y}y'+y'e^x+ye^x=0.

Solve for y', we obtain y'=-\dfrac{e^y+ye^x}{xe^y+e^x}

Plug (0, 1) into the right hand side, we get the slope of the tangent line m=y'(0)=-\dfrac{e+1}{1}=-(e+1).

So the tangent line is defined by y=-(e+1)x+1.

Exercise 7.24 Find the absolute maximum value of the function f(x)=x-e^{x}.

Solution.

The function f is differentiable over all real numbers. To find the absolute maximum, we study the monotonicity of the function.

The first derivative is f'(x)=1-e^x. Since e^x>1 for x>0 and e^x<1 for x<0, we know that f'(x)<0 for x>0 and f'(x)>0 for x<0.

The information shows that f(x) is increasing over (-\infty, 0) and decreasing over (0, \infty).

Thus, f(x) has a maximum f(0)=0-e^0=-1.

Exercise 7.25 Evaluate the integral \int e^{x} \sqrt{1+e^{x}} \mathrm{d} x.

Solution.

\begin{aligned} &\int e^{x} \sqrt{1+e^{x}} \mathrm{d} x\\ =& \int \sqrt{u} \mathrm{d} u \qquad\qquad u=1+e^x \\ =& \frac{2}{3} u^{\frac{3}{2}}+C\\ =& \frac{2}{3} (1+e^x)^{\frac{3}{2}}+C. \end{aligned}

Exercise 7.26 Evaluate the integral \int\left(e^{x}+e^{-x}\right)^{2} \mathrm{d} x.

Solution.

\begin{aligned} &\int\left(e^{x}+e^{-x}\right)^{2} \mathrm{d} x\\ =& \int (e^{2x}+2+e^{-2x}) \mathrm{d} x\\ =& \frac12 e^{2x}+2x-\frac12e^{-2x} + C. \end{aligned}

7.4 General Logarithmic and Exponential Functions

Exercise 7.27 Differentiate the function f(x)=x^{5}+5^{x}.

Solution.

f'(x)=5x+5^x\ln 5.

Exercise 7.28 Differentiate the function g(x)=x \sin \left(2^{x}\right).

Solution.

\begin{aligned} f'(x)=&\sin(2^x)+x\cos(2^x)(2^x)'\\ =&\sin(2^x)+x2^x\cos(2^x)\ln2. \end{aligned}

Exercise 7.29 Differentiate the function f(x)=\log _{5}\left(x e^{x}\right).

Solution.

\begin{aligned} f'(x)=&\frac{1}{\ln5}(\ln x+x)'\\ =&\frac{1}{\ln5}\left(\frac1x+1\right)\\ =&\frac{x+1}{x\ln5}. \end{aligned}

Exercise 7.30 Differentiate the function y=x^{\cos x}.

Solution.

\begin{aligned} f'(x)&=(e^{\cos x\ln x})'=x^{\cos x}(\cos x \ln x)'\\ &=x^{\cos x}\left(-\sin x\ln x+\frac{\cos x}{x}\right)\\ &=\frac{(\cos x-x\sin x\ln x)x^{\cos x}}{x}. \end{aligned}

Exercise 7.31 Evaluate the integral \int\left(x^{5}+5^{x}\right) \mathrm{d} x.

Solution.

\begin{aligned} &\int\left(x^{5}+5^{x}\right) \mathrm{d} x\\ =& \frac{x^6}{6}+\frac{5^x}{\ln 5}+C. \end{aligned}

Exercise 7.32 Evaluate the integral \int \frac{\log _{10} x}{x} \mathrm{d} x.

Solution.

\begin{aligned} &\int \frac{\log _{10} x}{x} \mathrm{d} x\\ =& \frac{1}{\ln 10}\int \frac{\ln x}{x} \mathrm{d} x\\ =& \frac{1}{\ln 10}\int \ln x ~\mathrm{d}(\ln x) \qquad u=\ln x\\ =& \frac{(\ln x)^2}{2\ln 10}+C. \end{aligned}

Exercise 7.33 Evaluate the integral \int_0^1 \frac{2^{x}}{2^{x}+1} \mathrm{d} x.

Solution.

Let u=2^x+1, then \mathrm{d}u=2^x\ln2\mathrm{d} x. Therefore, \begin{aligned} &\int \frac{2^{x}}{2^{x}+1} \mathrm{d} x\\ =&\frac{1}{\ln 2}\int \frac{1}{u} \mathrm{d} u\\ =&\frac{\ln u}{\ln 2} + C\\ =& \frac{\ln(2^x+1)}{\ln 2}+C. \end{aligned}

By FTC, \begin{aligned} \int_0^1 \frac{2^{x}}{2^{x}+1} \mathrm{d} x =&\frac{\ln(2^1+1)}{\ln 2}-\frac{\ln(2^0+1)}{\ln 2}\\ =&\frac{\ln3-\ln2}{\ln2}. \end{aligned}

Exercise 7.34 Find the area of the region bounded by the curves y=2^{x}, y=5^{x}, x=-1, and x=1

Solution.

Note that 5^x>2^x for x>0 and 2^x>5^x for x<0

So the area of the region is given by \begin{aligned} &\int_{-1}^0(2^x-5^x)\mathrm{d} x+\int_0^1(5^x-2^x)\mathrm{d}x\\ =&\left(\frac{2^x}{\ln 2}-\frac{5^x}{\ln 5}\right)\bigg|_{-1}^0+\left(\frac{5^x}{\ln 5}-\frac{2^x}{\ln 2}\right)\bigg|^{1}_0\\ =&\frac{\frac{16}{5}\ln 2-\frac12\ln5}{\ln 2\ln 5} \end{aligned}

7.5 Exponential Growth and Decay

Exercise 7.35 A sample of tritium-3 decayed to 94.5% of its original amount after a year.

1. What is the half-life of tritium-3?
2. How long would it take the sample to decay to 20% of its original amount?

Solution.

The decay of tritium-3 can be modeled by the function m(t)=m(0)e^{kt}.

We know that m(1)=0.945m(0) which implies that e^k=0.945. So k=\ln(0.945).

The half-life is the solution of the equation 0.5 = e^{t\ln(0.945)}. Taking \ln of both sides and solve for t, we get the half-life is \frac{\ln(0.5)}{\ln(0.945)}\approx 12.25 years.

Similarly, it takes the sample \frac{\ln(0.2)}{\ln(0.945)}\approx 28.45 years to decay to 20%.

Exercise 7.36 A curve passes through the point (0,5) and has the property that the slope of the curve at every point P is twice the y-coordinate of P. What is the equation of the curve?

Solution.

The slope of the line is given by the derivative y'. From the given information, we know that y'=2y which has the solution y(x)=Ce^{2x}.

Since the point (0,5) is on the curve, we have 5=y(0)=Ce^0. So C=5 and the equation of the curve is y=5e^{2x}.

Exercise 7.37 A freshly brewed cup of tea has temperature 97^\circC. The temperature of the room is 22^\circC. After 10 minutes, the temperature of the coffee is 47^\circC. Find the temperature of the coffee after 20 minutes. By Newton’s cooling law, the temperature of the coffee satisfies the equation y'(t)=k(y(t)-R), where R is the room temperature.

Solution.

Since R=22 and y(0)=97, by Newton’s cooling law, the temperature of the coffee can be modeled by y(t)=(98-23)e^{kt}+23=75e^{kt}+23.

We also know that y(10)=47 which gives us an equation 75e^{10k}+22=47.

Solve for k, we get t=-\frac{\ln 3}{10}.

Therefore, \begin{aligned} y(20)=&75e^{-2\ln 3}+22\\ =&75\cdot\left(e^{\ln 3}\right)^{-2}+22\\ =&\frac{91}{3}\approx 30.3^\circ \text{C}. \end{aligned}

Exercise 7.38 How long will it take an investment to double in value if the interest rate is 6% compounded continuously?

Solution.

The balance function of an investment with an annual rate 6% compounded continuous is given by B(t)=Pe^{0.06t}.

When the balance is doubled, it satisfies the equation 2P=Pe^{0.06}.

Solve the equation, we get that it takes about t=\ln2/0.06\approx 11.55 years that the investment will be doubled.

7.6 Inverse Trigonometric Functions

Exercise 7.39 Find the derivative of the function

y=\tan^{-1}\left(x^{2}\right)

Solution.

By the chain rule, we get y'=\dfrac{2x}{1+x^4}.

Exercise 7.40 Find the derivative of the function

y=\sin^{-1}(2x+1)

Solution.

By the chain rule, we get y'=\dfrac{2}{\sqrt{1-(2x+1)^2}}.

Exercise 7.41 Find the derivative of the function

f(\theta)=\arcsin\sqrt{\sin \theta}

Solution.

Apply the chain rule twice, we get y'=\dfrac{1}{\sqrt{1-\sin\theta}}\cdot\frac{1}{2\sqrt{\sin\theta}}\cdot\cos\theta.

Exercise 7.42 Find the derivative of the function

y=\arctan\sqrt{\frac{1-x}{1+x}}

Solution.

Note that \frac{1-x}{1+x}=\frac2{1+x}-1.

Apply the chain rule three times, we get \begin{aligned} y'=&\dfrac{1}{1+\frac{1-x}{1+x}}\cdot\frac{1}{2\sqrt{\frac{1-x}{1+x}}}\cdot(-\frac{2}{(x+1)^2})\\ =&-\frac{1}{2(x+1)}\sqrt{\frac{1+x}{1-x}}. \end{aligned}

Exercise 7.43 Evaluate the integral

\int_{1 / \sqrt{3}}^{\sqrt{3}} \frac{8}{1+x^{2}} \mathrm{d} x

Solution.

\begin{aligned} \int_{1/\sqrt{3}}^{\sqrt{3}} \frac{8}{1+x^{2}} \mathrm{d} x =&8\arctan(x)\big|_{1/\sqrt{3}}^{\sqrt{3}}\\ =&8\left(\frac{\pi}{3}-\frac{\pi}{6}\right)\\ =&\frac{4\pi}{3}. \end{aligned}

Exercise 7.44 Evaluate the integral

\int_{0}^{1 / 2} \frac{\sin ^{-1} x}{\sqrt{1-x^{2}}} \mathrm{d} x

Solution.

Let u=\sin^{-1}x. Then \begin{aligned} \int \frac{\sin ^{-1} x}{\sqrt{1-x^{2}}} \mathrm{d} x =\int u\mathrm{d} u=\frac12u^2+C=\frac12(\sin^{-1}x)^2+C. \end{aligned}

Therefore, \begin{aligned} &\int_{0}^{1 / 2} \frac{\sin ^{-1} x}{\sqrt{1-x^{2}}} \mathrm{d} x\\ =& \frac12(\sin^{-1}(1/2))^2-\frac12(\sin^{-1}0)^2\\ =&\frac12\left(\frac{\pi}{6}\right)^2=\frac{\pi}{72}. \end{aligned}

Exercise 7.45 Evaluate the integral

\int_{0}^{\pi / 2} \frac{\sin x}{1+\cos ^{2} x} \mathrm{d} x

Solution.

Let u=\cos x. Then \begin{aligned} &\int\frac{\sin x}{1+\cos ^{2} x} \mathrm{d} x\\ =&\int\frac{1}{1+u^2}\mathrm{d}u\\ =&\arctan(u)+C=\arctan(\cos x)+C. \end{aligned}

Therefore, \begin{aligned} &\int_{0}^{\pi / 2} \frac{\sin x}{1+\cos ^{2} x} \mathrm{d} x\\ =&\arctan(\cos(\pi/2))-\arctan(\cos 0)\\ =&\arctan(0)-\arctan(1)=-\frac{\pi}{4}. \end{aligned}

Exercise 7.46 Evaluate the integral

\int \frac{e^{2 x}}{\sqrt{1-e^{4 x}}} \mathrm{d} x

Solution.

Let u=e^{2x}. Then \mathrm{d}u=2e^{2x}\mathrm{d}x and \begin{aligned} &\int \frac{e^{2 x}}{\sqrt{1-e^{4 x}}} \mathrm{d} x\\ =&\frac12\int\frac{1}{\sqrt{1-u^2}}\mathrm{d}u\\ =&\frac12\arcsin(u)+C=\frac12\arcsin(e^{2x})+C. \end{aligned}

Exercise 7.47 Find the limit

\lim_{x \rightarrow \infty} \arccos \left(\frac{1+x^{2}}{1+2 x^{2}}\right)

Solution.

Because \cos x is continuous over its domain [-1, 1] and 0\le \frac{1+x^{2}}{1+2 x^{2}}\le 1.

Then \begin{aligned} \lim_{x\rightarrow \infty} \arccos \left(\frac{1+x^{2}}{1+2 x^{2}}\right) =&\arccos\left(\lim_{x \rightarrow \infty}\frac{1+x^{2}}{1+2 x^{2}}\right)\\ =&\arccos(\frac12)=\frac{\pi}{3}. \end{aligned}

Exercise 7.48 Find the limit

\lim_{x \rightarrow 0^{+}} \tan ^{-1}(\ln x)

Solution.

Because \lim\limits_{x \rightarrow 0^{+}}\ln x=-\infty and \lim\limits_{x \rightarrow -\infty} \tan^{-1}x=-\frac{\pi}{2}.

Then \lim_{x \rightarrow 0^{+}} \tan ^{-1}(\ln x)=-\frac{\pi}{2}.

7.7 L’hospital’s Rule

Exercise 7.49 Evaluate the limit if it exist. Otherwise, explain why.

\lim_{x \rightarrow 1} \frac{x^{3}-2 x^{2}+1}{x^{3}-1}

Solution.

Because \lim\limits_{x \rightarrow 1}(x^{3}-2 x^{2}+1)=0 and \lim_{x \rightarrow 1}(x^{3}-1)=0. Moreover, \begin{aligned} \lim_{x \rightarrow 1} \frac{(x^{3}-2 x^{2}+1)'}{(x^{3}-1)'} =& \lim_{x \rightarrow 1} \frac{3x^2-4x}{3x^2}\\ =&\frac{3-4}{3}=-\frac13. \end{aligned}

By L’Hospital’s rule, \lim_{x \rightarrow 1} \frac{x^{3}-2 x^{2}+1}{x^{3}-1}=\lim_{x \rightarrow 1} \frac{(x^{3}-2 x^{2}+1)'}{(x^{3}-1)'}=-\frac13.

Exercise 7.50 Evaluate the limit if it exist. Otherwise, explain why.

\lim_{x \rightarrow 0} \frac{\sin 4 x}{\tan 5 x}

Solution.

Since \lim\limits_{x \rightarrow 0}\sin 4 x=0 and \lim\limits_{x \rightarrow 0}\cos 4 x=0, and \lim_{x \rightarrow 0} \frac{(\sin 4 x)'}{(\tan 5 x)'}=\lim_{x \rightarrow 0} \frac{4\cos4x}{5\sec^2 5 x}=\frac45.

By L’Hospital’s rule, \lim_{x \rightarrow 0} \frac{\sin 4 x}{\tan 5 x}=\lim_{x \rightarrow 0} \frac{(\sin 4 x)'}{(\tan 5 x)'}=\frac45.

Exercise 7.51 Evaluate the limit if it exist. Otherwise, explain why.

\lim_{\theta \rightarrow \pi / 2} \frac{1-\sin \theta}{1+\cos 2 \theta}

Solution.

We will apply L’Hospital’s rule twice.

\begin{aligned} \lim_{\theta \rightarrow \pi / 2} \frac{1-\sin \theta}{1+\cos 2 \theta} =&\lim_{\theta \rightarrow \pi / 2} \frac{-\cos\theta}{-2\sin 2 \theta}\\ =&\lim_{\theta \rightarrow \pi / 2} \frac{-\sin\theta}{4\cos 2 \theta}\\ =&\frac{-1}{-4}=\frac14. \end{aligned}

Exercise 7.52 Evaluate the limit if it exist. Otherwise, explain why.

\lim_{x \rightarrow 0} \frac{\sin^{-1} x}{x}

Solution.

We may apply L’Hospital’s rule.

\begin{aligned} \lim_{x \rightarrow 0} \frac{\sin^{-1} x}{x} =&\lim_{\theta \rightarrow 0} \frac{1}{\sqrt{1-x^2}}\\ &=1. \end{aligned}

Exercise 7.53 Evaluate the limit if it exist. Otherwise, explain why.

\lim_{x \rightarrow \infty} x \sin(\pi / x)

Solution.

First write the expression as a quotient and then apply L’Hospital’s rule.

\begin{aligned} \lim_{x \rightarrow \infty} x \sin(\pi / x) =&\lim_{x \rightarrow \infty}\frac{\sin(\pi / x)}{1/x}\\ =&\lim_{x \rightarrow \infty}\frac{\cos(\pi / x)\cdot(\pi\cdot(1/x)')}{(1/x)'}\\ =&\lim_{x \rightarrow \infty}\pi\cos(\pi / x)=\pi. \end{aligned}

Exercise 7.54 Evaluate the limit if it exist. Otherwise, explain why.

\lim_{x \rightarrow 0^{+}}\left(\frac{1}{x}-\frac{1}{e^{x}-1}\right)

Solution.

First write the expression as a quotient and then apply L’Hospital’s rule twice.

\begin{aligned} \lim_{x \rightarrow 0^{+}}\left(\frac{1}{x}-\frac{1}{e^{x}-1}\right) =&\lim_{x \rightarrow 0^{+}}\frac{e^x-x-1}{x(e^x-1)}\\ =&\lim_{x \rightarrow 0^{+}}\frac{e^x-1}{(e^x-1)+xe^x}\\ =&\lim_{x \rightarrow 0^{+}}\frac{e^x}{2e^x+xe^x}=\frac12. \end{aligned}

Exercise 7.55 Evaluate the limit if it exist. Otherwise, explain why.

\lim_{x \rightarrow \infty}(x-\ln x)

Solution.

An estimate using the identity \ln(e^x)=x shows that the limit is \infty. To verify that, we factor out x. Note that \lim_{x \rightarrow \infty}\frac{\ln x}{x}=\lim_{x \rightarrow \infty}\frac{1}{x}=0.

Then \begin{aligned} \lim_{x \rightarrow \infty} (x-\ln x) =&\lim_{x \rightarrow \infty}x\left(1-\frac{\ln x}{x}\right)\\ =&\infty. \end{aligned}

Exercise 7.56 Find the limit

\lim_{x \rightarrow \infty} x^{1 / x}

Solution.

Using the identity \ln(e^x)=x, we may rewrite x^{1/x} as x^{1/x}=e^{\frac{\ln x}{x}}.

Since \lim_{x \rightarrow \infty}\frac{\ln x}{x}=\lim_{x \rightarrow \infty}\frac{1}{x}=0.

Then \lim_{x \rightarrow \infty} x^{1 / x}=e^{\lim\limits_{x \rightarrow \infty}\frac{\ln x}{x}}=e^0=1.