Exercises — L'Hôpital's rule — proof using linear approximation, 0 - 0, ∞ - ∞, other indeterminate forms

Type check. Top $\to \sin 0=0$, bottom $\to 0$. Form is $\frac{0}{0}$ ✔ — L'Hôpital is allowed. Differentiate top and bottom separately (not the quotient rule): $\frac{(\sin x{)}^{\prime }}{(x{)}^{\prime }}=\frac{\cos x}{1}.$ Take the limit: $\cos 0=1$. $\boxed{\underset{x\to 0}{\lim }\frac{\sin x}{x}=1.}$

Type check. $\tan 0=0$, bottom $\to 0$: form $\frac{0}{0}$ ✔. Recall $\frac{d}{dx}\tan x={\sec }^{2}x=\frac{1}{{\cos }^{2}x}$. $\frac{(\tan x{)}^{\prime }}{(x{)}^{\prime }}=\frac{{\sec }^{2}x}{1}\stackrel{x\to 0}{\to }\frac{1}{{\cos }^{2}0}=1.$ $\boxed{\underset{x\to 0}{\lim }\frac{\tan x}{x}=1.}$

Type check first! Top $\to \cos 0=1$, bottom $\to 0+1=1$. This is $\frac{1}{1}$ — not indeterminate. L'Hôpital is not allowed; the answer is just direct substitution: $\boxed{\underset{x\to 0}{\lim }\frac{\cos x}{x+1}=\frac{1}{1}=1.}$ Had you blindly differentiated, you'd get $\frac{-\sin x}{1}\to 0$ — the wrong answer. The form gate is not optional.

Type check. $e^0-1=0$, $\sin 0=0$: form $\frac{0}{0}$ ✔. Differentiate: top $(e^{2x}-1{)}^{\prime }=2e^{2x}$ (chain rule), bottom $(\sin 3x{)}^{\prime }=3\cos 3x$. $\frac{2e^{2x}}{3\cos 3x}\stackrel{x\to 0}{\to }\frac{2\cdot 1}{3\cdot 1}=\frac{2}{3}.$ $\boxed{\underset{x\to 0}{\lim }\frac{e^{2x}-1}{\sin 3x}=\frac{2}{3}.}$

Type check. $0-\sin 0=0$, $0^3=0$: $\frac{0}{0}$ ✔. First application: $\frac{1-\cos x}{3x^2}$. Still $\frac{0}{0}$ (top $\to 1-1=0$). Second application: $\frac{\sin x}{6x}$. Still $\frac{0}{0}$. Third application: $\frac{\cos x}{6}\to \frac{1}{6}$. $\boxed{\underset{x\to 0}{\lim }\frac{x-\sin x}{x^3}=\frac{1}{6}.}$ Why repeat? Each new fraction is again a legitimate $\frac{0}{0}$, so the rule re-applies. We stop the instant the top or bottom no longer both vanish.

Type check. As $x\to 0^{+}$: $x^2\to 0$ and $\ln x\to -\infty$, so form $0\cdot (-\infty )$ — indeterminate but not a fraction, so L'Hôpital can't touch it yet. Convert to a fraction. Put the $-\infty$ part on top: $x^2\ln x=\frac{\ln x}{1/x^2},\text{form }\frac{-\infty }{\infty }✔.$ Differentiate: top $(\ln x{)}^{\prime }=\frac{1}{x}$, bottom $(x^{-2}{)}^{\prime }=-2x^{-3}$. $\frac{1/x}{-2/x^3}=\frac{x^3}{-2x}=-\frac{x^2}{2}\stackrel{x\to 0^{+}}{\to }0.$ $\boxed{\underset{x\to 0^{+}}{\lim }{x}^2\ln x=0.}$ Why put $\ln x$ on top? Its derivative $1/x$ is simple; the alternative split $\frac{x^2}{1/\ln x}$ has an ugly derivative. Choose the split that simplifies.

Type check. As $x\to 0$, $\frac{1}{x}\to \infty$ and $\frac{1}{\sin x}\to \infty$: form $\infty -\infty$. Indeterminate — the two infinities may cancel to anything. Combine over a common denominator to expose a $\frac{0}{0}$: $\frac{1}{x}-\frac{1}{\sin x}=\frac{\sin x-x}{x\sin x},\text{form }\frac{0}{0}✔.$ First application: top $(\sin x-x{)}^{\prime }=\cos x-1$; bottom $(x\sin x{)}^{\prime }=\sin x+x\cos x$. $\frac{\cos x-1}{\sin x+x\cos x},\text{still }\frac{0}{0}.$ Second application: top $(\cos x-1{)}^{\prime }=-\sin x$; bottom $(\sin x+x\cos x{)}^{\prime }=\cos x+\cos x-x\sin x=2\cos x-x\sin x$. $\frac{-\sin x}{2\cos x-x\sin x}\stackrel{x\to 0}{\to }\frac{0}{2-0}=0.$ $\boxed{\underset{x\to 0}{\lim }\left(\frac{1}{x}-\frac{1}{\sin x}\right)=0.}$

Type check. $x^3\to \infty$, $e^x\to \infty$: form $\frac{\infty }{\infty }$ ✔. Apply three times (the polynomial degree drops by one each time; $e^x$ is untouched): $\frac{x^3}{e^x}\to \frac{3x^2}{e^x}\to \frac{6x}{e^x}\to \frac{6}{e^x}\stackrel{x\to \infty }{\to }0.$ $\boxed{\underset{x\to \infty }{\lim }\frac{x^3}{e^x}=0.}$ Interpretation. The exponential beats any fixed power — see Exponential & Logarithm Growth Rates. The picture below shows $e^x$ overtaking $x^3$ and never looking back.

Type check. As $x\to (\pi /2{)}^{-}$, $\cos x\to 0^{+}$, so $\sec x=\frac{1}{\cos x}\to +\infty$ and $\tan x=\frac{\sin x}{\cos x}\to +\infty$: form $\infty -\infty$. Combine: $\sec x-\tan x=\frac{1}{\cos x}-\frac{\sin x}{\cos x}=\frac{1-\sin x}{\cos x},\text{form }\frac{0}{0}✔.$ Apply once: top $(1-\sin x{)}^{\prime }=-\cos x$; bottom $(\cos x{)}^{\prime }=-\sin x$. $\frac{-\cos x}{-\sin x}=\frac{\cos x}{\sin x}\stackrel{x\to \pi /2}{\to }\frac{0}{1}=0.$ $\boxed{\underset{x\to (\pi /2{)}^{-}}{\lim }(\sec x-\tan x)=0.}$

Type check. Base $\to 1$, exponent $\to \infty$: form $1^{\infty }$ — indeterminate, and exponents are invisible to differentiation until we bring them down. Take a log. Let $y={\left(1+\frac{3}{x}\right)}^x$, so $\ln y=x\ln \left(1+\frac{3}{x}\right),\text{form }\infty \cdot 0.$ Convert to a fraction by putting the $\to 0$ factor on top over $1/x$: $\ln y=\frac{\ln (1+3/x)}{1/x},\text{form }\frac{0}{0}✔.$ Differentiate. Top: $\frac{d}{dx}\ln (1+3/x)=\frac{-3/x^2}{1+3/x}$. Bottom: $\frac{d}{dx}(1/x)=-1/x^2$. $\frac{-3/x^2}{1+3/x}\cdot \frac{1}{-1/x^2}=\frac{3}{1+3/x}\stackrel{x\to \infty }{\to }\frac{3}{1}=3.$ So $\ln y\to 3$, hence $y\to e^3$. $\boxed{\underset{x\to \infty }{\lim }{\left(1+\frac{3}{x}\right)}^x=e^3.}$

Type check. Base $x\to 0^{+}$, exponent $x\to 0$: form $0^0$ — indeterminate. Log it. Let $y=x^x$, so $\ln y=x\ln x$, form $0\cdot (-\infty )$. Convert: $\ln y=\frac{\ln x}{1/x}$, form $\frac{-\infty }{\infty }$ ✔. Differentiate: $\frac{1/x}{-1/x^2}=-x\stackrel{x\to 0^{+}}{\to }0$. So $\ln y\to 0$, hence $y\to e^0=1$. $\boxed{\underset{x\to 0^{+}}{\lim }{x}^x=1.}$

Type check. Base $\to \infty$, exponent $\to 0$: form ${\infty }^0$ — indeterminate. Log it. $y=x^{1/x}\Rightarrow \ln y=\frac{\ln x}{x}$, form $\frac{\infty }{\infty }$ ✔. Differentiate: $\frac{1/x}{1}=\frac{1}{x}\stackrel{x\to \infty }{\to }0$. So $\ln y\to 0$, hence $y\to e^0=1$. $\boxed{\underset{x\to \infty }{\lim }{x}^{1/x}=1.}$

Type check. Top $\to \infty$, bottom $\to \infty$: form $\frac{\infty }{\infty }$. But watch what L'Hôpital does: top ${}^{\prime }=e^x$; bottom ${}^{\prime }=\frac{2e^{2x}}{2\sqrt{e^{2x}+1}}=\frac{e^{2x}}{\sqrt{e^{2x}+1}}$. Ratio $=\frac{e^x\sqrt{e^{2x}+1}}{e^{2x}}=\frac{\sqrt{e^{2x}+1}}{e^x}$ — the fraction flipped and cycles forever. The RHS limit never resolves, so the rule gives nothing. Use algebra instead. Factor $e^{2x}$ out of the root: $\frac{e^x}{\sqrt{e^{2x}+1}}=\frac{e^x}{e^x\sqrt{1+e^{-2x}}}=\frac{1}{\sqrt{1+e^{-2x}}}\stackrel{x\to \infty }{\to }\frac{1}{\sqrt{1+0}}=1.$ $\boxed{\underset{x\to \infty }{\lim }\frac{e^x}{\sqrt{e^{2x}+1}}=1.}$ Lesson: L'Hôpital requires the RHS limit to exist. When it cycles, that hypothesis fails — retreat to algebra.

Direct (correct) evaluation. $\frac{x+\sin x}{x}=1+\frac{\sin x}{x}.$ Since $-1\le \sin x\le 1$, $\left|\frac{\sin x}{x}\right|\le \frac{1}{x}\to 0$ (squeeze). Hence $\boxed{\underset{x\to \infty }{\lim }\frac{x+\sin x}{x}=1.}$ What L'Hôpital would give. Top ${}^{\prime }=1+\cos x$, bottom ${}^{\prime }=1$, ratio $=1+\cos x$, which oscillates in $[0,2]$ and has no limit. No contradiction. L'Hôpital's theorem says: if $\lim f^{\prime }/g^{\prime }$ exists, then it equals $\lim f/g$. It says nothing when $\lim f^{\prime }/g^{\prime }$ fails to exist. The implication runs one way only — a non-existent $f^{\prime }/g^{\prime }$ limit does not condemn $f/g$. Here $f/g$ has a perfectly good limit, $1$.

The Taylor expansion of $\sin x$ near $0$ is $\sin x=x-\frac{x^3}{6}+\frac{x^5}{120}-\cdots .$ So $x-\sin x=\frac{x^3}{6}-\frac{x^5}{120}+\cdots$, and $\frac{x-\sin x}{x^3}=\frac{1}{6}-\frac{x^2}{120}+\cdots \stackrel{x\to 0}{\to }\frac{1}{6}.$ $\boxed{\underset{x\to 0}{\lim }\frac{x-\sin x}{x^3}=\frac{1}{6}.}$ Why they agree. L'Hôpital applied $k$ times to a $\frac{0}{0}$ isolates the ratio of the $k$-th order terms; that is exactly the ratio of leading nonzero Taylor coefficients. Here the numerator's first surviving term is $\frac{x^3}{6}$ and the denominator's is $x^3$, giving $\frac{1/6}{1}=\frac{1}{6}$ — the same $\frac{1}{6}$ we ground out with three derivative passes. This is L'Hôpital seen from orbit: compare the leading Taylor terms.

Determinate form ($\frac{\text{finite}}{\text{nonzero}}$)? ::: Just substitute — L'Hôpital forbidden. Product $0\cdot \infty$? ::: Rewrite as a fraction $f/(1/g)$, choose the split that simplifies. Difference $\infty -\infty$? ::: Common denominator to expose $\frac{0}{0}$. Power $0^0,{\infty }^0,1^{\infty }$? ::: Take $\ln$, solve, then exponentiate the result. Rule cycles or RHS oscillates? ::: Abandon it — use algebra, squeeze, or Taylor series.