06. 로피탈 정리L'Hôpital's Rule

Solution. Each limit is of the form $0/0$ as $x \to 0$, so we apply L'Hôpital's Rule.

(1) $\displaystyle\lim_{x \to 0} \frac{3x - \sin x}{x} = \lim_{x \to 0} \frac{3 - \cos x}{1} = 3 - 1 = 2$.

(2) $\displaystyle\lim_{x \to 0} \frac{\sqrt{1+x} - 1}{x} = \lim_{x \to 0} \frac{1/(2\sqrt{1+x})}{1} = \frac{1}{2}$.

(3) Let $f(x) = \sqrt{1+x} - 1 - x/2$ and $g(x) = x^2$. Then $f(0) = g(0) = 0$, and $f'(x) = \dfrac{1}{2\sqrt{1+x}} - \dfrac{1}{2}$, $g'(x) = 2x$ again give $f'(0) = g'(0) = 0$, so we apply L'Hôpital's Rule once more. Since $f''(x) = -\dfrac{1}{4}(1+x)^{-3/2}$ and $g''(x) = 2$, $$\lim_{x \to 0} \frac{\sqrt{1+x} - 1 - x/2}{x^2} = \lim_{x \to 0} \frac{f''(x)}{g''(x)} = \frac{-1/4}{2} = -\frac{1}{8}.$$

(4) Let $f(x) = x - \sin x$ and $g(x) = x^3$. Since $f'(x) = 1 - \cos x$, $g'(x) = 3x^2$ is again $0/0$, and $f''(x) = \sin x$, $g''(x) = 6x$ is again $0/0$, we apply L'Hôpital's Rule three times. Since $f'''(x) = \cos x$ and $g'''(x) = 6$, $$\lim_{x \to 0} \frac{x - \sin x}{x^3} = \lim_{x \to 0} \frac{\cos x}{6} = \frac{1}{6}.$$

Solution.

(1) As $x \to 0$ this is of the form $0/0$, so by L'Hôpital's Rule, $$\lim_{x \to 0} \frac{1 - \cos x}{x + x^2} = \lim_{x \to 0} \frac{\sin x}{1 + 2x} = \frac{0}{1} = 0.$$

(2), (3) Since $\sin x \to 0$ as $x \to 0$, this is of the form $0/0$. By L'Hôpital's Rule, $$\lim_{x \to 0} \frac{\sin x}{x^2} = \lim_{x \to 0} \frac{\cos x}{2x}.$$ As $x \to 0+$, the numerator $\cos x \to 1 \gt 0$ while the denominator $2x \to 0+$, so the limit is $+\infty$; as $x \to 0-$, the denominator $2x \to 0-$, so the limit is $-\infty$. Hence $$\lim_{x \to 0+} \frac{\sin x}{x^2} = +\infty, \qquad \lim_{x \to 0-} \frac{\sin x}{x^2} = -\infty.$$

Solution.

(1) As $x \to \pi/2$, both $\sec x \to \infty$ and $1 + \tan x \to \infty$, so this is of the form $\infty/\infty$. By L'Hôpital's Rule, $$\lim_{x \to \pi/2} \frac{\sec x}{1 + \tan x} = \lim_{x \to \pi/2} \frac{\sec x \tan x}{\sec^2 x} = \lim_{x \to \pi/2} \frac{\tan x}{\sec x} = \lim_{x \to \pi/2} \sin x = 1.$$

(2) As $x \to \infty$ this is of the form $\infty/\infty$, so by L'Hôpital's Rule, $$\lim_{x \to \infty} \frac{\ln x}{2\sqrt{x}} = \lim_{x \to \infty} \frac{1/x}{1/\sqrt{x}} = \lim_{x \to \infty} \frac{1}{\sqrt{x}} = 0.$$

(3) As $x \to \infty$ this is of the form $\infty/\infty$; applying L'Hôpital's Rule twice, $$\lim_{x \to \infty} \frac{e^x}{x^2} = \lim_{x \to \infty} \frac{e^x}{2x} = \lim_{x \to \infty} \frac{e^x}{2} = \infty.$$

Solution.

(1) Substituting $t = 1/x$, as $x \to \infty$ we have $t \to 0$, and the limit is of the form $0/0$, so by L'Hôpital's Rule, $$\lim_{x \to \infty} x \sin \frac{1}{x} = \lim_{t \to 0} \frac{\sin t}{t} = \lim_{t \to 0} \frac{\cos t}{1} = 1.$$

(2) This is of the form $0 \cdot (-\infty)$. Rewriting, $$\sqrt{x}\,\ln x = \frac{\ln x}{x^{-1/2}},$$ which is of the form $-\infty/\infty$ as $x \to 0+$. By L'Hôpital's Rule, $$\lim_{x \to 0+} \sqrt{x}\,\ln x = \lim_{x \to 0+} \frac{1/x}{-\frac12 x^{-3/2}} = \lim_{x \to 0+} (-2\sqrt{x}) = 0.$$

(3) This is of the form $\infty - \infty$. Combining the fractions, $$\frac{1}{\sin x} - \frac{1}{x} = \frac{x - \sin x}{x \sin x},$$ which is of the form $0/0$ as $x \to 0$. Applying L'Hôpital's Rule twice, $$\lim_{x \to 0} \frac{x - \sin x}{x \sin x} = \lim_{x \to 0} \frac{1 - \cos x}{\sin x + x\cos x} = \lim_{x \to 0} \frac{\sin x}{2\cos x - x \sin x} = \frac{0}{2} = 0.$$

Solution.

(1) This is of the form $1^{\infty}$. Let $y = (1+x)^{1/x}$, so $\ln y = \dfrac{\ln(1+x)}{x}$, which is of the form $0/0$ as $x \to 0+$. By L'Hôpital's Rule, $$\lim_{x \to 0+} \ln y = \lim_{x \to 0+} \frac{1/(1+x)}{1} = 1,$$ so $\displaystyle\lim_{x \to 0+} (1+x)^{1/x} = e^1 = e$.

(2) This is of the form $\infty^0$. Let $y = x^{1/x}$, so $\ln y = \dfrac{\ln x}{x}$, which is of the form $\infty/\infty$ as $x \to \infty$. By L'Hôpital's Rule, $$\lim_{x \to \infty} \ln y = \lim_{x \to \infty} \frac{1/x}{1} = 0,$$ so $\displaystyle\lim_{x \to \infty} x^{1/x} = e^0 = 1$.

Solution. From the Maclaurin expansion $\tan u = u + \dfrac{u^3}{3} + O(u^5)$ with $u = 2x$, $$\tan 2x = 2x + \frac{8x^3}{3} + O(x^5),$$ so $$\frac{\tan 2x}{x^3} = \frac{2}{x^2} + \frac{8}{3} + O(x^2).$$ Also $\sin bx = bx - \dfrac{b^3x^3}{6} + O(x^5)$, so $$\frac{\sin bx}{x} = b - \frac{b^3x^2}{6} + O(x^4).$$ Adding all three terms, $$\frac{\tan 2x}{x^3} + \frac{a}{x^2} + \frac{\sin bx}{x} = \frac{2+a}{x^2} + \left(\frac{8}{3} + b\right) + O(x^2).$$ For this to converge to $0$ as $x \to 0$, both the coefficient of $1/x^2$ and the constant term must vanish: $$2 + a = 0, \qquad \frac{8}{3} + b = 0.$$ Hence $a = -2$ and $b = -\dfrac{8}{3}$.

Solution.

(1) Substituting $t = 1/x$, as $x \to \pm\infty$ we have $t \to 0$, and the limit is of the form $0/0$, so by L'Hôpital's Rule, $$\lim_{x \to \pm\infty} x \tan \frac{1}{x} = \lim_{t \to 0} \frac{\tan t}{t} = \lim_{t \to 0} \frac{\sec^2 t}{1} = 1.$$ Hence the horizontal asymptote is $y = 1$ both as $x \to \infty$ and as $x \to -\infty$.

(2) As $x \to \infty$, both the numerator and denominator diverge to $\infty$; applying L'Hôpital's Rule twice, $$\lim_{x \to \infty} \frac{3x + e^{2x}}{2x + e^{3x}} = \lim_{x \to \infty} \frac{3 + 2e^{2x}}{2 + 3e^{3x}} = \lim_{x \to \infty} \frac{4e^{2x}}{9e^{3x}} = \lim_{x \to \infty} \frac{4}{9}e^{-x} = 0.$$ As $x \to -\infty$, $e^{2x} \to 0$ and $e^{3x} \to 0$, and since numerator and denominator both diverge to $-\infty$ we again apply L'Hôpital's Rule once to get $$\lim_{x \to -\infty} \frac{3x + e^{2x}}{2x + e^{3x}} = \lim_{x \to -\infty} \frac{3 + 2e^{2x}}{2 + 3e^{3x}} = \frac{3}{2}.$$ Hence the horizontal asymptote is $y = 0$ as $x \to \infty$, and $y = 3/2$ as $x \to -\infty$.

Solution.

(1) We find the limit of $y = (\sin x)^x$ as $x \to 0+$. Since $\ln y = x \ln \sin x$ is of the form $0 \cdot (-\infty)$, rewrite it as $$\ln y = \frac{\ln \sin x}{1/x},$$ which is of the form $-\infty/\infty$. By L'Hôpital's Rule, $$\lim_{x \to 0+} \ln y = \lim_{x \to 0+} \frac{\cot x}{-1/x^2} = \lim_{x \to 0+} \left(-x^2 \cot x\right) = \lim_{x \to 0+} \left(-x \cdot \frac{x}{\sin x} \cdot \cos x\right) = -0 \cdot 1 \cdot 1 = 0.$$ Hence $\displaystyle\lim_{x \to 0+}(\sin x)^x = e^0 = 1$, so $y = f(x)$ is continuous at $x = 0$ exactly when $a = 1$.

(2) For $x \in (0, \pi)$ we have $0 \lt \sin x \leq 1$, so $\ln \sin x \leq 0$; since $x \gt 0$, $$\ln f(x) = x \ln \sin x \leq 0,$$ i.e., $f(x) \leq 1$. Equality holds when $\ln \sin x = 0$, i.e., $\sin x = 1$, which occurs at $x = \pi/2$ (where $f(\pi/2) = 1^{\pi/2} = 1$). Meanwhile at $x = \pi$ we have $f(\pi) = 0^{\pi} = 0 \lt 1$, and by part (1), $f(0) = a = 1$. Hence the maximum value of $y = f(x)$ on $[0, \pi]$ is $1$, attained at both $x = 0$ and $x = \pi/2$.

Solution. By Cauchy's Mean Value Theorem, there exists $c \in (a, b)$ such that $\dfrac{f(b) - f(a)}{g(b) - g(a)} = \dfrac{f'(c)}{g'(c)}$.

(1) For $f(x) = x$, $g(x) = x^2$, $(a, b) = (-2, 0)$, $$\frac{f(0) - f(-2)}{g(0) - g(-2)} = \frac{0 - (-2)}{0 - 4} = -\frac{1}{2},$$ and $f'(c) = 1$, $g'(c) = 2c$, so $$\frac{1}{2c} = -\frac{1}{2}$$ gives $c = -1$. Since $c = -1 \in (-2, 0)$, this satisfies the condition.

(2) For $f(x) = x^3/3 - 4x$, $g(x) = x^2$, $(a, b) = (0, 3)$, $$\frac{f(3) - f(0)}{g(3) - g(0)} = \frac{(9 - 12) - 0}{9 - 0} = -\frac{1}{3},$$ and $f'(c) = c^2 - 4$, $g'(c) = 2c$, so $$\frac{c^2 - 4}{2c} = -\frac{1}{3}$$ gives $3c^2 + 2c - 12 = 0$, i.e., $$c = \frac{-2 \pm \sqrt{4 + 144}}{6} = \frac{-1 \pm \sqrt{37}}{3}.$$ Of these, only $c = \dfrac{-1 + \sqrt{37}}{3}$ ($\approx 1.69$) lies in $(0, 3)$.

Proof.

(1) Let $v = X(b) - X(a) = (f(b) - f(a),\, g(b) - g(a))$, and define $$\varphi(t) = f(t)\bigl(g(b) - g(a)\bigr) - g(t)\bigl(f(b) - f(a)\bigr).$$ Then $\varphi$ is continuous on $[a, b]$ and differentiable on $(a, b)$. Computing, $$\varphi(a) = f(a)g(b) - f(a)g(a) - g(a)f(b) + g(a)f(a) = f(a)g(b) - g(a)f(b),$$ $$\varphi(b) = f(b)g(b) - f(b)g(a) - g(b)f(b) + g(b)f(a) = f(a)g(b) - g(a)f(b),$$ so $\varphi(a) = \varphi(b)$. By Rolle's Theorem, there exists $c \in (a, b)$ with $\varphi'(c) = 0$, i.e., $$\varphi'(c) = f'(c)\bigl(g(b) - g(a)\bigr) - g'(c)\bigl(f(b) - f(a)\bigr) = 0.$$ This means the (scalar) cross product of $X'(c) = (f'(c), g'(c))$ and $v = X(b) - X(a)$ is $0$, so $X'(c)$ is parallel to $v$.

(2) Consider $f(t) = \cos t$, $g(t) = \sin t$, $h(t) = t$, $(a, b) = (0, 2\pi)$, which are continuous on $[0, 2\pi]$ and differentiable on $(0, 2\pi)$. Then $$X(2\pi) - X(0) = (\cos 2\pi - \cos 0,\ \sin 2\pi - \sin 0,\ 2\pi - 0) = (0, 0, 2\pi),$$ so for $X'(c) = (-\sin c, \cos c, 1)$ to be parallel to $(0, 0, 2\pi)$ we would need $$-\sin c = 0 \quad \text{and} \quad \cos c = 0$$ simultaneously. But since $\sin^2 c + \cos^2 c = 1$, no $c$ satisfies $\sin c = \cos c = 0$ at once. Hence no $c \in (a, b)$ makes $X'(c)$ parallel to $X(b) - X(a)$ in this case, so for space curves such a $c$ does not always exist.

Solution. We examine the limit of the ratio of the two functions as $x \to \infty$.

(1) $\displaystyle\lim_{x \to \infty} \frac{e^x}{x^2}$ is of the form $\infty/\infty$; applying L'Hôpital's Rule twice, $$\lim_{x \to \infty} \frac{e^x}{x^2} = \lim_{x \to \infty} \frac{e^x}{2x} = \lim_{x \to \infty} \frac{e^x}{2} = \infty.$$ Hence $f(x) = e^x$ grows faster than $g(x) = x^2$.

(2) $\displaystyle\lim_{x \to \infty} \frac{3^x}{2^x} = \lim_{x \to \infty} \left(\frac{3}{2}\right)^x = \infty$ since $3/2 \gt 1$. Hence $f(x) = 3^x$ grows faster than $g(x) = 2^x$.

(3) $\displaystyle\lim_{x \to \infty} \frac{x^2}{\ln x}$ is of the form $\infty/\infty$, so by L'Hôpital's Rule, $$\lim_{x \to \infty} \frac{x^2}{\ln x} = \lim_{x \to \infty} \frac{2x}{1/x} = \lim_{x \to \infty} 2x^2 = \infty.$$ Hence $f(x) = x^2$ grows faster than $g(x) = \ln x$.

(4) $\displaystyle\lim_{x \to \infty} \frac{\ln x}{x^{1/n}}$ is of the form $\infty/\infty$, so by L'Hôpital's Rule, $$\lim_{x \to \infty} \frac{\ln x}{x^{1/n}} = \lim_{x \to \infty} \frac{1/x}{\frac{1}{n}x^{1/n - 1}} = \lim_{x \to \infty} n x^{-1/n} = 0.$$ Hence $g(x) = x^{1/n}$ grows faster than $f(x) = \ln x$ (that is, a logarithm grows slower than any positive power of $x$).

(5) Since $a \gt b \gt 0$, we have $a/b \gt 1$, and $$\lim_{x \to \infty} \frac{a^x}{b^x} = \lim_{x \to \infty} \left(\frac{a}{b}\right)^x = \infty.$$ Hence $f(x) = a^x$ grows faster than $g(x) = b^x$.

(6) Changing base, $\log_a x = \dfrac{\ln x}{\ln a}$ and $\log_b x = \dfrac{\ln x}{\ln b}$, so $$\frac{f(x)}{g(x)} = \frac{\log_a x}{\log_b x} = \frac{\ln b}{\ln a}$$ is a constant independent of $x$ (since $a \gt b \gt 1$, $\ln a \gt \ln b \gt 0$, and this constant lies between $0$ and $1$). Hence $f(x) = \log_a x$ and $g(x) = \log_b x$ have the same growth rate, related proportionally by $\log_b x = \dfrac{\ln a}{\ln b}\log_a x$.

Solution.

(1) As $x \to \infty$, $\sqrt{x^2+5} = x\sqrt{1 + 5/x^2}$, and $g(x) = (2\sqrt{x}-1)^2 = 4x - 4\sqrt{x} + 1$. $$\lim_{x \to \infty} \frac{f(x)}{g(x)} = \lim_{x \to \infty} \frac{x\sqrt{1+5/x^2}}{4x - 4\sqrt{x}+1} = \lim_{x \to \infty} \frac{\sqrt{1+5/x^2}}{4 - 4/\sqrt{x} + 1/x} = \frac{1}{4},$$ a nonzero finite constant. Hence $f(x) = \sqrt{x^2+5}$ and $g(x) = (2\sqrt{x}-1)^2$ have the same growth rate ($g$ is consistently about $4$ times $f$).

(2) Since $\ln(x+999) = \ln x + \ln\!\left(1 + \dfrac{999}{x}\right)$, and this is of the form $\infty/\infty$ as $x \to \infty$, by L'Hôpital's Rule, $$\lim_{x \to \infty} \frac{\ln(x+999)}{\ln x} = \lim_{x \to \infty} \frac{1/(x+999)}{1/x} = \lim_{x \to \infty} \frac{x}{x+999} = 1.$$ Hence $f(x) = \ln(x+999)$ and $g(x) = \ln x$ have the same growth rate, and are in fact asymptotically equal.

(3) First, $f(x) = \log_2(x^2) = \dfrac{2\ln x}{\ln 2}$ and $g(x) = \log_{10}(10x) = 1 + \dfrac{\ln x}{\ln 10} \sim \dfrac{\ln x}{\ln 10}$, so $$\lim_{x \to \infty} \frac{f(x)}{g(x)} = \lim_{x \to \infty} \frac{2\ln x/\ln 2}{\ln x/\ln 10} = \frac{2\ln 10}{\ln 2},$$ a nonzero finite constant. Hence $f, g$ have the same growth rate (both are constant multiples of $\ln x$).

Meanwhile $h(x) = 1/\sqrt{x} \to 0$ and $r(x) = 1/x^2 \to 0$, and $$\lim_{x \to \infty} \frac{r(x)}{h(x)} = \lim_{x \to \infty} \frac{1/x^2}{1/\sqrt{x}} = \lim_{x \to \infty} x^{-3/2} = 0,$$ so $r(x)$ converges to $0$ faster than $h(x)$; that is, $h(x)$ decreases more slowly than $r(x)$. In summary, $f, g$ have the same growth rate (as they diverge), while $h, r$ both converge to $0$ with $h$ decreasing more slowly than $r$.

Solution. We take the limit to be $x \to \infty$ throughout. Recall $f = o(g)$ means $\displaystyle\lim \frac{f(x)}{g(x)} = 0$, and $f = O(g)$ means $\left|\dfrac{f(x)}{g(x)}\right|$ stays bounded for $x$ sufficiently large.

(1) Since $\displaystyle\frac{x}{\ln x} \to \infty$ ($x$ grows faster than $\ln x$), $x = o(\ln x)$ is false; since the ratio is unbounded, $x = O(\ln x)$ is also false.

(2) Since $\displaystyle\frac{x^2}{x^3+1} \to 0$, $x^2 = o(x^3+1)$ is true. Since $o$ implies $O$, $x^2 = O(x^3+1)$ is also true.

(3) Since $\displaystyle\frac{x+\sin x}{x} = 1 + \frac{\sin x}{x} \to 1 \neq 0$, $x + \sin x = o(x)$ is false. But since this ratio converges to $1$ (hence is bounded), $x + \sin x = O(x)$ is true.

(4) Since $\displaystyle\frac{e^x+x^2}{e^x} = 1 + \frac{x^2}{e^x} \to 1 \neq 0$, $e^x+x^2 = o(e^x)$ is false. Since the ratio converges to $1$, $e^x+x^2 = O(e^x)$ is true.

(5) Since $\displaystyle\frac{x}{e^x} \to 0$, $x = o(e^x)$ is true, and hence $x = O(e^x)$ is also true.

Solution. As before, we take the limit to be $x \to \infty$.

(1) $\displaystyle\frac{x}{2x} = \frac{1}{2}$ is a constant for all $x$, hence bounded. So $x = O(2x)$ is true.

(2) $\displaystyle\frac{e^x}{e^{2x}} = e^{-x} \to 0$, so $e^x = o(e^{2x})$ is true.

(3) $\displaystyle\frac{1/(x+3)}{1/x} = \frac{x}{x+3} \to 1$, which is bounded, so $\dfrac{1}{x+3} = O\!\left(\dfrac{1}{x}\right)$ is true.

(4) As $x \to \infty$, $\ln(x^2+1) \sim \ln(x^2) = 2\ln x$, so $$\lim_{x \to \infty} \frac{\ln x}{\ln(x^2+1)} = \lim_{x \to \infty} \frac{\ln x}{2\ln x + \ln(1+1/x^2)} = \frac{1}{2}.$$ Since this limit is $1/2 \neq 0$, $\ln x = o(\ln(x^2+1))$ is false.