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$$\begin{aligned} \lim_{x\rightarrow0}\underbrace{\frac{1-\cos x}{\sin^2 x}}_{\genfrac{}{}{0in}{}{\mathrm{num}\to0}{\mathrm{den}\to0}} &= \lim_{x\rightarrow0}\frac{\sin x}{2\sin x\cos x} =\lim_{x\rightarrow0}\frac{1}{2\cos x} ={\frac{1}{2}} \end{aligned}$$

If we plug in $x=0$, the numerator is zero, and the denominator is

$\sec 0 = \dfrac{1}{\cos 0}=\dfrac{1}{1}=1$. So the limit is $\dfrac{0}{1}=0$.

Be careful: you cannot use l'Hopital's Rule here, because the fraction does not give an indeterminate form. If you try to differentiate the numerator and the denominator, you get an expression whose limit does not exist:

$\displaystyle \lim_{x \to 0}\dfrac{1}{\sec x \tan x}=\displaystyle \lim_{x \to 0}\cos x\cdot \dfrac{\cos x}{\sin x}=DNE$.

If we plug $x=0$ into the denominator, we get 1. However, the numerator is an indeterminate form: $\tan 0 =0$, while $\displaystyle \lim_{x \to 0^+}\csc x=\infty$ and $\displaystyle \lim_{x \to 0^-}\csc x=-\infty$. If we use $\csc x = \frac{1}{\sin x}$, our expression becomes

$$\begin{aligned} \lim_{x\to0}\frac{\tan x\cdot (x^2+5)}{\sin x \cdot e^x}& \\ \text{Since plugging in $x=0$ makes both the numerator and the denominator equal to zero, this is a candidate for l'H\^ospital's Rule. However, a much easier way is to simplify the trig first.} \\ \lim_{x\to0}\frac{\tan x\cdot (x^2+5)}{\sin x \cdot e^x}&= \lim_{x\to0}\frac{\sin x\cdot (x^2+5)}{\cos x \cdot\sin x \cdot e^x}\\&= \lim_{x\to0}\frac{x^2+5}{\cos x \cdot e^x}\\ &=\frac{0^2+5}{\cos(0)\cdot e^0}=5 \end{aligned}$$

$\displaystyle \lim_{x \to \infty}\sqrt{2x^2+1}-\sqrt{x^2+x}$ has the indeterminate form $\infty - \infty$. To get a better idea of what's going on, let's rationalize.

$$\begin{aligned} \lim_{x \to \infty}\sqrt{2x^2+1}-\sqrt{x^2+x}&= \lim_{x \to \infty}\left(\sqrt{2x^2+1}-\sqrt{x^2+x}\right)\left( \frac{\sqrt{2x^2+1}+\sqrt{x^2+x}}{\sqrt{2x^2+1}+\sqrt{x^2+x}}\right)\\ &=\lim_{x \to \infty} \frac{(2x^2+1)-(x^2+x)}{\sqrt{2x^2+1}+\sqrt{x^2+x}}= \lim_{x \to \infty} \frac{x^2-x+1}{\sqrt{2x^2+1}+\sqrt{x^2+x}} \\ \text{Here, we have the indeterminate form $\frac{\infty}{\infty}$, so l'H\^opital's Rule applies. However, if we try to use it here, we quickly get a huge mess. Instead, remember how we dealt with these kinds of limits in the past: factor out the highest power of the denominator, which is $x$.} \\ &=\lim_{x \to \infty} \frac{x\left(x-1+\frac{1}{x}\right)}{\sqrt{x^2(2+\frac{1}{x^2})}+\sqrt{x^2(1+\frac{1}{x})}}\\ &=\lim_{x \to \infty} \frac{x\left(x-1+\frac{1}{x}\right)}{x\left(\sqrt{2+\frac{1}{x^2}}+\sqrt{1+\frac{1}{x}}\right)}\\ &=\lim_{x \to \infty} \underbrace{\frac{x-1+\frac{1}{x}}{\sqrt{2+\frac{1}{x^2}}+\sqrt{1+\frac{1}{x}}}}_{\genfrac{}{}{0in}{}{\mathrm{num}\to \infty}{\mathrm{den}\to \sqrt{2}+1}}\\ &=\infty \end{aligned}$$

If we plug in $x=0$, both numerator and denominator become zero. So, we have exactly one of the indeterminate forms that l'Hopital's Rule applies to.

$$\begin{aligned} \lim_{x\rightarrow 0}\underbrace{\dfrac{\sin(x^3+3x^2)}{\sin^2x}}_{\genfrac{}{}{0in}{}{\mathrm{num}\to0}{\mathrm{den}\to 0}} &= \lim_{x\rightarrow 0}\dfrac{(3x^2+6x)\cos(x^3+3x^2)}{2\sin x\cos x} \\ \text{If we plug in $x=0$, still we find that both the numerator and the denominator go to zero. We could jump in with another iteration of l'H\^opital's Rule. However, the derivatives would be a little messy, so we use limit laws and break up the fraction into the product of two fractions. If both limits exist:} \\ \lim_{x\rightarrow 0}\dfrac{(3x^2+6x)\cos(x^3+3x^2)}{2\sin x\cos x}&=\left(\lim_{x\rightarrow 0}\dfrac{x^2+2x}{\sin x}\right)\cdot \left(\lim_{x\rightarrow 0}\dfrac{3\cos(x^3+3x^2)}{2\cos x}\right) \\ \text{We can evaluate the right-hand limit by simply plugging in $x=0$:} \\ &=\dfrac{3}{2}\lim_{x\rightarrow 0}\underbrace{\dfrac{x^2+2x}{\sin x}}_ {\genfrac{}{}{0in}{}{\mathrm{num}\to0}{\mathrm{den}\to 0}} \\ &= \dfrac{3}{2}\lim_{x\rightarrow 0}\dfrac{2x+2}{\cos x}\\ &=\frac{3}{2}\left(\frac{2}{1}\right)=3\end{aligned}$$

$$\lim_{x\rightarrow1}\frac{\log(x^3)}{x^2-1} =\lim_{x\rightarrow1}\underbrace{\frac{3\log(x)}{x^2-1}}_{\genfrac{}{}{0in}{}{\mathrm{num}\to0}{\mathrm{den}\to0}} =\lim_{x\rightarrow1}\frac{3/x}{2x} =\frac{3}{2}$$

• Solution 1. $$\lim_{x\rightarrow 0}\frac{e^{-1/x^2}}{x^4} =\lim_{x\rightarrow 0}\underbrace{\frac{\frac{1}{x^4}}{e^{1/x^2}}}_{\genfrac{}{}{0in}{}{\mathrm{num}\to\infty}{\mathrm{den}\to\infty}} =\lim_{x\rightarrow 0}\frac{\frac{-4}{x^5}}{\frac{-2}{x^3}e^{1/x^2}} =\lim_{x\rightarrow 0}\underbrace{\frac{\frac{2}{x^2}}{e^{1/x^2}}}_{\genfrac{}{}{0in}{}{\mathrm{num}\to\infty}{\mathrm{den}\to\infty}} =\lim_{x\rightarrow 0}\frac{\frac{-4}{x^3}}{\frac{-2}{x^3}e^{1/x^2}} =\lim_{x\rightarrow 0}\frac{2}{e^{1/x^2}} ={0}$$ since, as $x\rightarrow 0$, the exponent $\frac{1}{x^2}\rightarrow\infty$ so that $e^{1/x^2}\rightarrow\infty$ and $e^{-1/x^2}\rightarrow 0$.
• Solution 2. $$\lim_{x\rightarrow 0}\frac{e^{-1/x^2}}{x^4} =\lim_{t=\frac{1}{x^2}\rightarrow \infty}\frac{e^{-t}}{t^{-2}} =\lim_{t\rightarrow \infty}\underbrace{\frac{t^2}{e^{t}}}_{\genfrac{}{}{0in}{}{\mathrm{num}\to\infty}{\mathrm{den}\to\infty}} =\lim_{t\rightarrow \infty}\underbrace{\frac{2t}{e^{t}}}_{\genfrac{}{}{0in}{}{\mathrm{num}\to\infty}{\mathrm{den}\to\infty}} =\lim_{t\rightarrow \infty}\frac{2}{e^{t}} ={0}$$

$$\lim_{x\rightarrow 0}\underbrace{\frac{xe^x}{\tan (3x)}}_{\genfrac{}{}{0in}{}{\mathrm{num}\to0}{\mathrm{den}\to0}} = \lim_{x\rightarrow 0}\frac{e^x+xe^x}{3\sec^2 (3x)} =\frac{1}{3}$$

$\displaystyle \lim_{x \to 0} \sin^2 x = 0$, and $\displaystyle \lim_{x \to 0}\frac{1}{x^2}=\infty$, so we have the form $0^\infty$. (Note that $\sin^2 x$ is positive, so our root is defined.) This is not an indeterminate form: $\lim\limits_{x \to 0}\sqrt[x^2]{\sin^2 x}=0$.

$\displaystyle \lim_{x \to 0} \cos x =1$ and $\displaystyle \lim_{x \to 0}\frac{1}{x^2}=\infty$, so $\displaystyle \lim_{x \to 0}(\cos x)^{\frac{1}{x^2}}$ has the indeterminate form $1^\infty$. We want to use l'Hopital, but we need to get our function into a fractional indeterminate form. So, we'll use a logarithm.

$$\begin{aligned} y:&=(\cos x)^{\frac{1}{x^2}}\\ \log y &= \log \left((\cos x)^{\frac{1}{x^2}}\right)=\frac{1}{x^2}\log(\cos x)=\frac{\log \cos x}{x^2}\\ \lim_{x \to 0}\log y &=\lim_{x \to 0}\underbrace{\frac{\log \cos x}{x^2}}_{\genfrac{}{}{0in}{}{\mathrm{num}\to0}{\mathrm{den}\to 0}} = \lim_{x \to 0} \frac{\frac{-\sin x}{\cos x}}{2x}= \lim_{x \to 0} \underbrace{\frac{-\tan x}{2x}}_{\genfrac{}{}{0in}{}{\mathrm{num}\to0}{\mathrm{den}\to 0}} = \lim_{x \to 0}\frac{-\sec^2x}{2}\\ &=\lim_{x \to 0}\frac{-1}{2\cos^2 x}=-\frac{1}{2}\\ \text{Therefore, } \lim_{x \to 0} y &=\lim_{x \to 0}e^{\log y}=e^{-1/2}=\frac{1}{\sqrt{e}} \end{aligned}$$

• Solution 1

$$\begin{aligned} y:&= e^{x \log x} = (e^x)^{\log x} \\ \lim_{x \to 0^+} y &=\lim_{x \to 0^+} (e^x)^{\log x} \\ \text{This has the form $1^{-\infty} = \frac{1}{1^\infty}$, and $1^{\infty}$ is an indeterminate form. We want to use l'H\^opital, but we need to get a different indeterminate form. So, we'll use logarithms.} \\ \lim_{x \to 0^+} \log y &=\lim_{x \to 0 ^+} \log\left((e^x)^{\log x} \right) =\lim_{x \to 0 ^+} \log x \log\left(e^x \right)=\lim_{x \to 0 ^+} (\log x)\cdot x \\ \text{This has the indeterminate form $0 \cdot \infty$, so we need one last adjustment before we can use l'H\^opital's Rule.} \\ &=\lim_{x \to 0 ^+} \underbrace{\frac{\log x}{\frac{1}{x}}}_{\genfrac{}{}{0in}{}{\mathrm{num} \to -\infty}{\mathrm{den} \to \infty}} =\lim_{x \to 0 ^+}\frac{\frac{1}{x}}{\frac{-1}{x^2}} =\lim_{x \to 0 ^+}-x=0 \\ \text{Now, we can figure out what happens to our original function, $y$:} \\ \lim_{x \to 0^+} y &=\lim_{x \to 0^+} e^{\log y} = e^0=1 \end{aligned}$$

• Solution 2

$$\begin{aligned} y:&=e^{x\log x}=\left(e^{\log x}\right)^x=x^x\\ \lim_{x \to 0^+} y &=\lim_{x \to 0^+}x^x \\ \text{We have the indeterminate form $0^0$. We want to use l'H\^opital, but we need a different indeterminate form. So, we'll use logarithms.} \\ \lim_{x \to 0^+}\log y &=\lim_{x \to 0^+}\log(x^x)=\lim_{x \to 0^+}x\log x \\ \text{Now we have the indeterminate form $0 \cdot \infty$, so we need one last adjustment before we can use l'H\^opital's Rule.} \\ \lim_{x \to 0^+} y &=\lim_{x \to 0^+}\underbrace{\frac{\log x}{\frac{1}{x}}}_{\genfrac{}{}{0in}{}{\mathrm{num}\to 0}{\mathrm{den} \to -\infty}} =\lim_{x \to 0^+}\frac{\frac{1}{x}}{\frac{-1}{x^2}} =\lim_{x \to 0^+}-x=0 \\ \text{Now, we can figure out what happens to our original function, $y$:} \\ \lim_{x \to 0^+} y &=\lim_{x \to 0^+} e^{\log y} = e^0=1 \end{aligned}$$

First, note that the function exists near 0: $x^2$ is positive, so $\log(x^2)$ exists; near 0, $\log x^2$ is negative, so $-\log(x^2)$ is positive, so $\left[-\log(x^2)\right]^x$ exists even when $x$ is negative.

Since $\displaystyle \lim_{x \rightarrow 0} -\log(x^2)=\infty$ and $\displaystyle \lim_{x \rightarrow 0}x=0$, we have the indeterminate form $\infty^0$. We need l'Hopital, but we need to manipulate our function into an appropriate form. We do this using logarithms.

$$\begin{aligned} y:&=\left[-\log(x^2)\right]^x\\ \log y &= \log \left( \left[-\log(x^2)\right]^x\right)= \underbrace{x}_{\to 0}\cdot\underbrace{\log\left(\underbrace{-\log(x^2)}_{\to\infty} \right)}_{\to\infty}=\frac{\log\left(-\log(x^2)\right)}{\frac{1}{x}}\\ \lim_{x \to 0}\log y &=\lim_{x \to 0} \underbrace{\frac{\log\left(-\log(x^2)\right)}{\frac{1}{x}}}_{\genfrac{}{}{0in}{}{\mathrm{num}\to \infty}{\mathrm{den}\to \pm\infty}} =\lim_{x \to 0} \frac{\frac{-\frac{2}{x}}{-\log(x^2)}}{\frac{-1}{x^2}} =\lim_{x \to 0} \underbrace{\frac{-2x}{\log(x^2)}}_{\genfrac{}{}{0in}{}{\mathrm{num}\to0}{\mathrm{den}\to-\infty}}=0 \\ \text{Now, we're ready to figure out our original limit.} \\ \lim_{x\to 0} y &= \lim_{x \to 0} e^{\log y}=e^0=1 \end{aligned}$$