Exercises
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Exercises
Evaluate the following limits.
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$\lim\limits_{x\to \infty} x\tan(1/x) $Explorations
- Show that $\lim\limits_{x\to\infty}(1+1/x)^x = e$.
- Evaluate $\lim\limits_{x\to\infty}(a+1/x)^x$, $\lim\limits_{x\to\infty}(1+b/x)^x$ and $\lim\limits_{x\to\infty}(1+1/x)^{cx}$.
- Compound interest describes a situation where a single payment ($P$) is made into an account which then earns a fixed interest rate ($r$) a certain number of times per year ($n$) for a fixed number of years ($t$). In this scenario, the final value of the account is given by $A=P(1+r/n)^{nt}$. Find a formula for the future value of the account as $n\to\infty$, i.e. when the interest is compound infinitely many times or "continuously" over the investment period.
- In 1696, the Marquis de l'Hospital published the method we're disussing here In his book, Analyse des Infiniment Petits. The example that he used was $$\lim\limits_{x\to a}\frac{\sqrt{2a^3x-x^4}-a\sqrt[3]{aax}}{a-\sqrt[4]{ax^3}}$$ Assuming $a>0$ evaluate this limit.
- What value of $a$ makes the function $$f(x) = \begin{cases}\frac{1+\cos x - \sin x}{x}, & x \ne 0 \\ a, & x=0 \\ \end{cases}$$
- Why does l'Hospital's Rule not help with the limit $\lim\limits_{x\to\infty}\frac{\sqrt{2x+1}}{\sqrt{x+1}}$? Evaluate the limit.
- Why does l'Hospital's Rule not help with the limit $\lim\limits_{x\to 0^+}\frac{\csc x}{\cot x}$? Evaluate the limit.
Icons courtesy of icons8.com
Evaluate $\lim\limits_{x\to 1} \frac{x-1}{x^2-1}$.
Evaluating the expression at 1 gives $\frac{0}{0}$ so l'Hospital's Rule applies.
$$\lim\limits_{x\to 1} \frac{x-1}{x^2-1}$$ $$\lim\limits_{x\to 1} \frac{1}{2x}$$ $$\frac{1}{2 \cdot 1}$$ $$\frac{1}{2}$$Evaluate $\lim\limits_{x\to 0} \frac{3^x-4^x}{x}$.
Evaluating the expression at 0 gives $\frac{0}{0}$ so l'Hospital's Rule applies.
$$\lim\limits_{x\to 0} \frac{3^x-4^x}{x}$$ $$\lim\limits_{x\to 0} \frac{3^x \ln (3) -4^x \ln(4) }{1}$$ $$\frac{3^0 \ln (3) -4^0 \ln(4) }{1}$$ $$\ln (3) - \ln(4)$$Evaluate $\lim\limits_{x\to 1} \frac{x^8-1}{x^{12}-1}$.
Evaluating the expression at 1 gives $\frac{0}{0}$ so l'Hospital's Rule applies.
$$\lim\limits_{x\to 1} \frac{x^8-1}{x^{12}-1}$$ $$\lim\limits_{x\to 1} \frac{8x^7}{12x^{11}}$$ $$\lim\limits_{x\to 1} \frac{2}{3x^{4}}$$ $$\frac{2}{3 \cdot 1^4}$$ $$\frac{2}{3}$$Evaluate $\lim\limits_{x\to 2} \frac{\sqrt{2x}-2}{x-2}$.
Evaluating the expression at 2 gives $\frac{0}{0}$ so l'Hospital's Rule applies.
$$\lim\limits_{x\to 2} \frac{\sqrt{2x}-2}{x-2}$$ $$\lim\limits_{x\to 2} \frac{\frac{2}{2\sqrt{2x}}}{1}$$ $$\frac{1}{\sqrt{2\cdot 2}}$$ $$\frac{1}{4}$$Evaluate $\lim\limits_{x\to 0} \frac{x 3^x}{2^x-1}$.
Evaluating the expression at 0 gives $\frac{0}{0}$ so l'Hospital's Rule applies.
$$\lim\limits_{x\to 0} \frac{x 3^x}{2^x-1}$$ $$\lim\limits_{x\to 0} \frac{1 \cdot 3^x + x 3^x \ln 3}{2^x \ln 2}$$ $$\frac{3^0 + 0 \cdot 3^0 \ln 3}{2^0 \ln 2}$$ $$\frac{1}{\ln 2}$$Evaluate $\lim\limits_{x\to \pi/2} (x-\pi/2)\tan x $.
Evaluating the expression at $\pi/2$ gives $0 \cdot \infty$ so l'Hospital's Rule applies if we maniuplate the expression first.
$$\lim\limits_{x\to \pi/2} (x-\pi/2)\tan x $$ $$\lim\limits_{x\to \pi/2} \frac{x-\pi/2}{\frac{1}{\tan x}} $$ $$\lim\limits_{x\to \pi/2} \frac{x-\pi/2}{\cot x} $$Now the expression evaluates to $0/0$ so we can apply l'Hospital's rule.
$$\lim\limits_{x\to \pi/2} \frac{1}{-\csc^2 x} $$ $$-\frac{1}{\csc^2 \frac{\pi}{2}}$$ $$-1$$Evaluate $\lim\limits_{x\to \infty} \sqrt{x} e^{-x}$.
Evaluating the expression at $\infty$ gives $\infty \cdot 0$ so l'Hospital's Rule applies if we maniuplate the expression first.
$$\lim\limits_{x\to \infty} \sqrt{x} e^{-x}$$ $$\lim\limits_{x\to \infty} \frac{\sqrt{x}}{e^{x}}$$Now the limit evaluates to $\infty / \infty$ so l'Hospital's Rule applies.
$$\lim\limits_{x\to \infty} \frac{\frac{1}{2\sqrt{x}}}{e^{x}}$$ $$\lim\limits_{x\to \infty} \frac{1}{2e^{x}\sqrt{x}}$$ $$0$$Evaluate $\lim\limits_{x\to \infty} x\tan(1/x) $.
Evaluating the expression at $\infty$ gives $\infty \cdot 0$ so l'Hospital's Rule applies if we maniuplate the expression first.
$$\lim\limits_{x\to \infty} x\tan(1/x) $$ $$\lim\limits_{x\to \infty} \frac{\tan(1/x)}{1/x} $$I'm going to take a shortcut here by letting $u=1/x$ then I'll apply l'Hospital's Rule.
$$\lim\limits_{x\to \infty} \frac{\tan(u)}{u} $$ $$\lim\limits_{x\to \infty} \frac{\sec^2 u \frac{du}{dx}}{1 \cdot \frac{du}{dx}} $$ $$\lim\limits_{x\to \infty} \frac{\sec^2 u}{1} $$Do you see why I made the substitution now? The whole calculation got cleaner since I never had to actually include the derivative of the $1/x$ part. Now I'll switch back to the $x$ variable.
$$\lim\limits_{x\to \infty} \sec^2 \frac{1}{x} $$ $$\sec^2 0 $$ $$1$$Evaluate $\lim\limits_{x\to \infty} (x + \ln x) $.
Both expressions, $x$ and $\ln x$ are going to infinity so l'Hospital's Rule doesn't apply here and we have
$$\lim\limits_{x\to \infty} (x + \ln x) = \infty$$Evaluate $\lim\limits_{x\to 1^+} (\log(x^3-1) - \ln(x^3-1))$.
Try first applying the Change of Base formula.
Evaluate $\lim\limits_{x\to 0^+} (\ln x - \ln (\sin x))$.
Both functions are going to $+\infty$ so the question is at what rate are the going? If the rates are similar then this may still be a finite result.
$$\begin{aligned} \lim\limits_{x\to 0^+} (\ln x - \ln (\sin x)) &= \lim\limits_{x\to 0^+} \ln \frac{x}{\sin x} \\ &= \ln \lim\limits_{x\to 0^+} \frac{x}{\sin x} \\ &= \ln \lim\limits_{x\to 0^+} \frac{1}{\cos x} \\ &= \ln \frac{1}{1} \\ &= \ln 1 \\ &= 0 \end{aligned}$$Evaluate $\lim\limits_{x\to 0} (\sin x)^x $.
This expression evaluates to $0^0$ so we can put into into an indeterminate form by looking at its logarithmically.
$$y = \lim\limits_{x\to 0} (\sin x)^x $$ $$\ln y = \ln \lim\limits_{x\to 0} (\sin x)^x $$ $$\ln y = \lim\limits_{x\to 0} \ln (\sin x)^x $$ $$\ln y = \lim\limits_{x\to 0} x \ln (\sin x) $$Now the limit evaluates to $0 \cdot \infty$ so we can maniuplate it again to get it to be an indeterminate form. $$\ln y = \lim\limits_{x\to 0} \frac{ \ln (\sin x) }{\frac{1}{x}}$$ $$\ln y = \lim\limits_{x\to 0} \frac{ \frac{\cos x}{\sin x} }{\frac{-1}{x^2}}$$ $$\ln y = \lim\limits_{x\to 0} \frac{ \cot x }{\frac{-1}{x^2}}$$
This is still an indeterminate form, $\infty / \infty$, so we'll apply l'Hospital's Rule again.
$$\ln y = \lim\limits_{x\to 0} \frac{ -\csc^2 x }{\frac{2}{x^3}}$$ $$\ln y = \lim\limits_{x\to 0} \frac{ -x^3\csc^2 x }{2}$$ $$\ln y = -\frac{ 0^3\cdot 0}{2}$$ $$\ln y = 0$$ $$y = \lim\limits_{x\to 0} (\sin x)^x = e^0 = 1$$Evaluate $\lim\limits_{x\to 0} (e^x+x)^{1/x}$.
This expression evaluates to $1^\infty$ so we can put into into an indeterminate form by looking at its logarithmically.
$$y = \lim\limits_{x\to 0} (e^x+x)^{1/x}$$ $$\ln y = \lim\limits_{x\to 0} \ln (e^x+x)^{1/x}$$ $$\ln y = \lim\limits_{x\to 0} \frac{1}{x} \ln (e^x+x)$$ $$\ln y = \lim\limits_{x\to 0} \frac{\ln (e^x+x)}{x}$$ $$\ln y = \lim\limits_{x\to 0} \frac{\frac{e^x+1}{e^x+x}}{1}$$ $$\ln y = \frac{e^0+1}{e^0+0}$$ $$\ln y = 2$$ $$y = \lim\limits_{x\to 0} (e^x+x)^{1/x} = e^2$$Show that $\lim\limits_{x\to\infty}(1+1/x)^x = e$.
$$y=\lim\limits_{x\to\infty}(1+1/x)^x$$ $$\ln y = \lim\limits_{x\to\infty}\ln (1+1/x)^x$$ $$\ln y = \lim\limits_{x\to\infty}x \ln (1+1/x)$$ $$\ln y = \lim\limits_{x\to\infty} \frac{\ln (1+1/x)}{\frac{1}{x}}$$ $$\ln y = \lim\limits_{x\to\infty} \frac{\frac{\frac{dy}{dx}}{1+1/x}}{\frac{dy}{dx}}$$ $$\ln y = \lim\limits_{x\to\infty} \frac{1}{1+1/x}$$ $$\ln y = \frac{1}{1+0}$$ $$\ln y = 1$$ $$y = \lim\limits_{x\to\infty}(1+1/x)^x = e^1=e$$
What value of $a$ makes the function $$f(x) = \begin{cases}\frac{1+\cos x - \sin x}{x}, & x \ne 0 \\ a, & x=0 \\ \end{cases}$$
$$a = \lim\limits_{x\to 0} \frac{1+\cos x - \sin x}{x} = -1$$
Why does l'Hospital's Rule not help with the limit $\lim\limits_{x\to\infty}\frac{\sqrt{2x+1}}{\sqrt{x+1}}$? Evaluate the limit.
$$\lim\limits_{x\to\infty}\frac{\sqrt{2x+1}}{\sqrt{x+1}} = \sqrt{2}$$
Why does l'Hospital's Rule not help with the limit $\lim\limits_{x\to 0^+}\frac{\csc x}{\cot x}$? Evaluate the limit.
Both the numerator and denominator are approaching $+\infty$ as $x\to 0^+$ so l'Hospital's Rule applies here. If we do that we get
$$\begin{aligned} \lim\limits_{x\to 0^+}\frac{\csc x}{\cot x} &= \lim\limits_{x\to 0^+}\frac{-\csc x \cot x}{-\csc^2 x} \\ &= \lim\limits_{x\to 0^+}\frac{\cot x}{\csc x} \end{aligned}$$Notice how the numerator and denominator just flipped so this is still an indeterminate form. If we apply l'Hospital's Rule again, the two parts of the fraction will flip again, taking us back to where we started and begining a never ending loop.
A better approach would be just to simplify the expression using trigonometry definitions.
$$\begin{aligned} \lim\limits_{x\to 0^+}\frac{\csc x}{\cot x} &= \lim\limits_{x\to 0^+}\frac{\frac{1}{\sin x}}{\frac{\cos x}{\sin x}} \\ &= \lim\limits_{x\to 0^+} \frac{1}{\sin x} \cdot \frac{\sin x}{\cos x} \\ &= \lim\limits_{x\to 0^+} \frac{1}{\cos x} \\ &= 1 \end{aligned}$$