L'Hôpital's Rule & Indeterminate Forms

Top: $e^0 - 1 = 0$. Bottom: $\sin 0 = 0$. Form $\dfrac{0}{0}$.

$\dfrac{(e^{3x} - 1)'}{(\sin 2x)'} = \dfrac{3e^{3x}}{2\cos 2x}$.

At $x = 0$: $\dfrac{3 \cdot 1}{2 \cdot 1} = \dfrac{3}{2}$ — not indeterminate.

The limit is $\dfrac{3}{2}$.

$\log x \to \infty$ and $\sqrt{x} \to \infty$. Form $\dfrac{\infty}{\infty}$.

$\dfrac{(\log x)'}{(\sqrt{x})'} = \dfrac{1/x}{1/(2\sqrt{x})}$.

Simplify: $\dfrac{2\sqrt{x}}{x} = \dfrac{2}{\sqrt{x}} \to 0$ as $x \to \infty$.

The limit is $0$.

Top: $0 - 0 = 0$. Bottom: $0$. Form $\dfrac{0}{0}$.

Round 1: $\dfrac{\cos x - x\sin x - \cos x}{3x^2} = \dfrac{-x\sin x}{3x^2}$.

At $0$: $\dfrac{0}{0}$ again. Round 2: $\dfrac{-\sin x - x\cos x}{6x}$, still $\dfrac{0}{0}$. Round 3: $\dfrac{-2\cos x + x\sin x}{6} \to \dfrac{-2}{6}$.

The limit is $-\dfrac{1}{3}$.

$x^2 \to 0$ and $\log x \to -\infty$. Form $0 \cdot (-\infty)$.

Keep the log on top: $x^2\log x = \dfrac{\log x}{x^{-2}}$.

$\dfrac{-\infty}{\infty}$ — L'Hôpital applies.

$\dfrac{1/x}{-2x^{-3}} = -\dfrac{x^3}{2x} = -\dfrac{x^2}{2}$.

$-\dfrac{x^2}{2} \to 0$.

$x \to \infty$ and $\sin\left(\dfrac{2}{x}\right) \to \sin 0 = 0$. Form $\infty \cdot 0$.

Move $x$ down: $\dfrac{\sin(2/x)}{1/x}$.

$\dfrac{0}{0}$ — L'Hôpital applies.

$\dfrac{\cos(2/x)\cdot(-2/x^2)}{-1/x^2} = 2\cos\left(\dfrac{2}{x}\right)$.

$2\cos 0 = 2$.

$x - \dfrac{\pi}{2} \to 0$ and $\tan x \to \infty$. Form $0 \cdot \infty$.

Move $\tan x$ down as $\cot x$: $\dfrac{x - \pi/2}{\cot x}$.

$\cot\dfrac{\pi}{2} = 0$, so $\dfrac{0}{0}$.

$\dfrac{1}{-\csc^2 x} = -\sin^2 x$.

$-\sin^2\dfrac{\pi}{2} = -1$.

Both terms $\to \infty$. Form $\infty - \infty$.

$\dfrac{(e^x - 1) - x}{x(e^x - 1)}$.

Top $\to 0$, bottom $\to 0$: $\dfrac{0}{0}$.

Round 1: $\dfrac{e^x - 1}{e^x - 1 + xe^x}$, still $\dfrac{0}{0}$. Round 2: $\dfrac{e^x}{2e^x + xe^x}$.

$\dfrac{1}{2 + 0} = \dfrac{1}{2}$.

$\dfrac{1}{x} \to \infty$ and $\dfrac{1}{\log(1 + x)} \to \infty$. Form $\infty - \infty$.

$\dfrac{\log(1 + x) - x}{x\log(1 + x)}$.

$\dfrac{0}{0}$.

Round 1: $\dfrac{\frac{1}{1+x} - 1}{\log(1 + x) + \frac{x}{1+x}}$, still $\dfrac{0}{0}$. Round 2: $\dfrac{-\frac{1}{(1+x)^2}}{\frac{1}{1+x} + \frac{1}{(1+x)^2}}$.

At $0$: $\dfrac{-1}{1 + 1} = -\dfrac{1}{2}$.

$\sec x \to \infty$ and $\tan x \to \infty$. Form $\infty - \infty$.

$\dfrac{1}{\cos x} - \dfrac{\sin x}{\cos x} = \dfrac{1 - \sin x}{\cos x}$.

$\dfrac{1 - 1}{0}$: $\dfrac{0}{0}$.

$\dfrac{-\cos x}{-\sin x} = \dfrac{\cos x}{\sin x}$.

$\dfrac{0}{1} = 0$.

Base $\to 1$, exponent $\to \infty$: $1^\infty$.

$\log y = 2x\log\left(1 + \dfrac{3}{x}\right)$.

$\infty \cdot 0$; move $x$ down: $\log y = \dfrac{2\log(1 + 3/x)}{1/x}$, a $\dfrac{0}{0}$.

$\dfrac{2 \cdot \frac{-3/x^2}{1 + 3/x}}{-1/x^2} = \dfrac{6}{1 + 3/x} \to 6$, so $L = 6$.

$\lim y = e^{6}$.

Base $\to 0^+$, exponent $\to 0$: $0^0$.

$\log y = x\log x$.

$0 \cdot (-\infty)$; keep the log on top: $\dfrac{\log x}{1/x}$, a $\dfrac{-\infty}{\infty}$.

$\dfrac{1/x}{-1/x^2} = -x \to 0$, so $L = 0$.

$\lim y = e^{0} = 1$.

Base $\to \infty$, exponent $\to 0$: $\infty^0$.

$\log y = \dfrac{\log(1 + x^2)}{\log x}$.

Already a fraction: $\dfrac{\infty}{\infty}$.

$\dfrac{2x/(1 + x^2)}{1/x} = \dfrac{2x^2}{1 + x^2} \to 2$, so $L = 2$.

$\lim y = e^{2}$.

Base $\to 1$, exponent $\to \infty$: $1^\infty$.

$\log y = \dfrac{\log(\cos 2x)}{x^2}$.

Already a fraction: $\dfrac{0}{0}$.

Round 1: $\dfrac{-2\sin 2x/\cos 2x}{2x} = \dfrac{-\tan 2x}{x}$, still $\dfrac{0}{0}$. Round 2: $\dfrac{-2\sec^2 2x}{1} \to -2$, so $L = -2$.

$\lim y = e^{-2}$.