Topic 7 Inverse Functions
7.1 Inverse Functions - Definition and Properties
Exercise 7.1 If \(f(x)=x^{5}+x^{3}+x,\) find \(f^{-1}(3)\) and \(\left(f^{-1}\right)'(3)\).
Solution.
Since \(f(f^{-1}(x))=x\), by change rule, we have \[ (f^{-1})'(x)=\dfrac{1}{f'(f^{-1}(x))} \]
To find \(f'\left(f^{-1}(3)\right)\), we first find \(f'\): \[ f'(x)=(x^{5}+x^{3}+x)'=5x^4+3x^2+1 \] 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 \[ (f^{-1})'(3)=\dfrac{1}{f'(f^{-1}(3))}=\dfrac{1}{f'(1)}=\dfrac{1}{5+3+1}=\dfrac{1}{9}. \]
Exercise 7.2 Find a formula for the inverse of the function \(f(x)=\dfrac{4 x-1}{2 x+3}\).
Solution.
Let \(y=f^{-1}(x)\). By the definition of \(f^{-1}(x)\), we have \[ x=f(f^{-1}(x))=f(y)=\dfrac{4 y-1}{2 y+3}. \]
Solve for \(y\) from this equation, we get \[ y=f^{-1}(x)=-\dfrac{3x+1}{2x-4}. \]
Exercise 7.3 Find a formula for the inverse of the function \(f(x)=1+\sqrt{2+3x}\).
Solution.
To get the inverse function, solve for \(y\) from the following equation \[ x=1+\sqrt{2+3y}, \] we get \[ f^{-1}(x)=y=\frac13(x-1)^2-\frac23. \]
Exercise 7.4 Find an explicit formula for \(f^{-1}\) where \(f(x)=x^{4}+1, \quad x \ge 0\).
Solution.
Solve for \(y\) from the following equation \[ x=y^4+1, \] we get \[ f^{-1}(x)=y=\sqrt[4]{x-1}. \]
Remark: Since the domain of \(f\) is \(x\ge 0\), the range of \(f^{-1}\) is \(y\ge 0\). That’s why we take \(y=\sqrt[4]{x-1}\).
Exercise 7.5 Find \(\left(f^{-1}\right)^{\prime}(a)\), where \(f(x)=2 x^{3}+3 x^{2}+7 x+4, \quad a=4\).
Solution.
Solve \(f(x)=4\), we get \(f^{-1}(4)=x=0\). As the derivative of \(f\) is \(f'(x)=6x^2+3x+7\), apply the formula \[ (f^{-1})'(x)=\dfrac{1}{f'(f^{-1}(x))}, \] we get \[ (f^{-1})'(4)=\dfrac{1}{f'(f^{-1}(4))}=\dfrac{1}{f'(0)}=\dfrac{1}{7}. \]
Exercise 7.6 Find \(\left(f^{-1}\right)^{\prime}(a)\), where \(f(x)=x^{3}+3 \sin x+2 \cos x, \quad a=2\).
Solution.
Solve \(f(x)=2\), we get \(f^{-1}(2)=x=0\). As the derivative of \(f\) is \(f'(x)=3x^2+3\cos x-2\sin x\), apply the formula \[ (f^{-1})'(x)=\dfrac{1}{f'(f^{-1}(x))}, \] we get \[ (f^{-1})'(2)=\dfrac{1}{f'(f^{-1}(2))}=\dfrac{1}{f'(0)}=\dfrac{1}{3}. \]
Exercise 7.7 If \(g\) is an increasing function such that \(g(2)=8\) and \(g^{\prime}(2)=5,\) calculate \(\left(g^{-1}\right)^{\prime}(8)\).
Solution.
Since \(g(2)=8\), \(g^{-1}(8)=2\). By the formula \[ (f^{-1})'(x)=\dfrac{1}{f'(f^{-1}(x))}, \] we get \[ (g^{-1})'(8)=\dfrac{1}{g'(g^{-1}(8))}=\dfrac{1}{g'(2)}=\dfrac{1}{5}. \]
Exercise 7.8 If \(f(x)=\int_{3}^{x} \sqrt{1+t^{3}} d t,\) find \(\left(f^{-1}\right)^{\prime}(0)\).
Solution.
By Fundamental Theorem of Calculus, we know that \[ f'(x)=\sqrt{1+x^3}\ge 0 \] over its domain \(x\ge -1\). Then \(f(x)\) is increasing over \([-1, \infty]\). By Fundamental Theorem of Calculus, \(f(3)=\int_3^3\sqrt{1+t^3}\mathrm{d} t=0\). So \(f^{-1}(0)=3\).
Therefore, \[ (f^{-1})'(0)=\dfrac{1}{f'(f^{-1}(0))}=\dfrac{1}{f'(3)}=\dfrac{1}{\sqrt{28}}=\frac{\sqrt{7}}{14}. \]
7.2 The Natural Logarithmic Function
Exercise 7.9 Find the limit \[ \lim_{x \to \infty}[\ln (2+x)-\ln (1+x)]. \]
Solution.
\[ \begin{aligned} &\lim_{x \to \infty}[\ln (2+x)-\ln (1+x)]\\ =&\lim_{x \to \infty}\ln\left(\dfrac{2+x}{1+x}\right)\\ =&\ln\left(\lim_{x \to \infty}\dfrac{2+x}{1+x}\right)\\ =&\ln 1=0\\ \end{aligned} \]
Remark: In the second last step, we used the trick that \[ \dfrac{2+x}{1+x}=\dfrac{\frac2x+1}{\frac1x+1}. \]
Exercise 7.10 Differentiate the function \[ f(x)=\sqrt{x} \ln x. \]
Solution.
Apply the chain rule, we get \[ f'(x)=(\sqrt{x} \ln x)'=\frac{\ln x}{2\sqrt{x}}+\frac{\sqrt{x}}{x}=\frac{\sqrt{x}(\ln x+2)}{2x}. \]
Exercise 7.11 Differentiate the function \[ f(x)=\ln \left(\sin ^{2} x\right). \]
Solution.
Apply the chain rule twice, we get \[ f'(x)=\dfrac{1}{\sin^2x}(\sin ^{2} x)'=\dfrac{1}{\sin^2x}(2\sin x)(\sin x)'=2\tan x. \]
Exercise 7.12 Differentiate the function \[ y=\frac{1}{\ln x}. \]
Solution.
Apply the chain rule, we get \[ f'(x)=-\dfrac{1}{(\ln x)^2}(\ln x)'=-\dfrac{1}{x(\ln x)^2}. \]
Exercise 7.13 Differentiate the function \[ y=(\ln \tan x)^{2}. \]
Solution.
Apply the chain rule twice, we get \[ f'(x)=2(\ln \tan x)(\ln \tan x)'=\dfrac{2(\ln \tan x)}{\tan x}(\tan x)'=\dfrac{2\sec^2 x(\ln \tan x)}{\tan x}. \]
Exercise 7.14 Find \(f^{\prime\prime}(e)\), where \[ f(x)=\frac{\ln x}{x}. \]
Solution.
Apply the quotient rule and simplify, we get \[ f'(x)=\dfrac{1- \ln x}{x^2} \]
Apply the quotient rule again and simplify, we get \[ f''(x)=\dfrac{2\ln x-3}{x^3} \]
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.
- What is the half-life of tritium-3?
- 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\(^\circ\)C. The temperature of the room is 22\(^\circ\)C. After 10 minutes, the temperature of the coffee is 47\(^\circ\)C. 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 \[ \lim_{x \rightarrow \infty} x^{1 / x}=e^{\lim\limits_{x \rightarrow \infty}\frac{\ln x}{x}}=e^0=1. \]