$$\lim_{x\to 0} \frac{\sin x}{x} \to \frac{0}{0}$$ so we can use L’Hôpital’s Rule. \begin{align*} \lim_{x\to 0} \frac{\sin x}{x} &= \lim_{x\to 0} \frac{\frac{d}{dx}(\sin x)}{\frac{d}{dx}(x)} \\ \\ &= \lim_{x\to 0} \frac{\cos x}{1} \\ \\ &= \frac{1}{1} = 1 \quad \cmark \end{align*}

$$\lim_{x\to 0} \frac{\cos x – 1}{x} \to \frac{1 – 1}{0} \to \frac{0}{0}$$ so we can use L’Hôpital’s Rule. \begin{align*} \lim_{x\to 0} \frac{\cos x – 1}{x} &= \lim_{x\to 0} \frac{\frac{d}{dx}(\cos x – 1)}{\frac{d}{dx}(x)} \\ \\ &= \lim_{x\to 0} \frac{-\sin x}{1} \\ \\ &= \frac{0}{1} = 0 \quad \cmark \end{align*}

$$\lim_{x\to \infty} \frac{3x^2 -5x + 2}{4x^2 -6} \to \frac{\infty}{\infty}$$ so we can use L’Hôpital’s Rule. \begin{align*} \lim_{x\to \infty} \frac{3x^2 -5x + 2}{4x^2 -6} &= \lim_{x\to \infty} \frac{\frac{d}{dx}(3x^2 -5x + 2)}{\frac{d}{dx}(4x^2 -6)} \\ \\ &= \lim_{x\to \infty} \frac{6x -5}{8x} \\ \\ \end{align*} This is also in the form of $\dfrac{\infty}{\infty}$, and so we can apply L’Hôpital’s Rule again:
\begin{align*} \phantom{\lim_{x\to \infty} \frac{3x^2 -5x + 2}{4x^2 -6}} &= \lim_{x\to \infty} \frac{\frac{d}{dx}(6x -5)}{\frac{d}{dx}(8x)} \\ \\ &= \lim_{x\to \infty} \frac{6}{8} \\ \\ &= \frac{6}{8} = \frac{3}{4} \quad \cmark \end{align*}

$$\lim_{x\to \infty} \frac{x^2}{(\ln x)^2} \to \frac{\infty}{\infty}$$ so we can use L’Hôpital’s Rule. \begin{align*} \lim_{x\to \infty} \frac{x^2}{(\ln x)^2} &= \lim_{x\to \infty} \frac{\frac{d}{dx}(x^2)}{\frac{d}{dx}\left[(\ln x)^2\right]} \\ \\ &= \lim_{x\to \infty} \frac{2x}{(2\ln x)\cdot (\frac{1}{x})} \\ \\ &= \lim_{x\to \infty} \frac{2x^2}{(2\ln x)} \end{align*} The last line is again in the form $\dfrac{\infty}{\infty}$, so let’s again apply L’Hôpital’s Rule:
\begin{align*} \phantom{\lim_{x\to \infty} \frac{x^2}{(\ln x)^2}} &= \lim_{x\to \infty} \frac{\frac{d}{dx}(2x^2)}{\frac{d}{dx}(2\ln x)} \\ \\ &= \lim_{x\to \infty} \frac{4x}{2 \cdot (\frac{1}{x})} \\ \\ &= \lim_{x\to \infty} 2x^2 \\ \\ &= \infty \quad \cmark \end{align*}

$$\lim_{x\to 2} \frac{x^2 – 4}{x+ 2} = \frac{0}{4} = 0 \quad \cmark$$ That’s the answer; we cannot use L’Hôpital’s Rule, since this is not in the form of either $\dfrac{0}{0}$ or $\dfrac{\infty}{\infty}$.
(Remember to always check! Using it incorrectly here would lead to a wrong answer.)

$$\lim_{x\to 0} \frac{x}{\tan x} \to \frac{0}{0}$$ so we can use L’Hôpital’s Rule. \begin{align*} \lim_{x\to 0} \frac{x}{\tan x} &= \lim_{x\to 0} \frac{\frac{d}{dx}(x)}{\frac{d}{dx}(\tan x)} \\ \\ &= \lim_{x\to 0} \frac{1}{\sec^2 x} \\ \\ &= \frac{1}{1} = 1 \quad \cmark \end{align*}

$$\lim_{x\to 0} \frac{x}{e^x} = \frac{0}{1} = 0 \quad \cmark$$ That’s the answer; we CANNOT use L’Hôpital’s Rule, since this is not in the form of either $\dfrac{0}{0}$ or $\dfrac{\infty}{\infty}$.
(Remember to always check! Using it incorrectly here would lead to a wrong answer.)

$$\lim_{x\to 1} \frac{\ln x}{x-1} \to \frac{0}{0}$$ so we can use L’Hôpital’s Rule. \begin{align*} \lim_{x\to 1} \frac{\ln x}{x-1} &= \lim_{x\to 1} \frac{\frac{d}{dx}(\ln x)}{\frac{d}{dx}(x-1)} \\ \\ &= \lim_{x\to 1} \frac{1/x}{1} \\ \\ &= 1 \quad \cmark \end{align*}

$$\lim_{x\to 0} \frac{5^x – 2^x}{x} \to \frac{1-1}{0} \to \frac{0}{0}$$ so we can use L’Hôpital’s Rule. \begin{align*} \lim_{x\to 0} \frac{5^x – 2^x}{x} &= \lim_{x\to 0} \frac{\frac{d}{dx}(5^x – 2^x)}{\frac{d}{dx}(x)} \\ \\ &=\lim_{x\to 0} \frac{\ln 5 \cdot 5^x – \ln2 \cdot 2^x}{1} \\ \\ &= \frac{\ln 5 \cdot 1 – \ln2 \cdot 1}{1} \\ \\ &= \ln 5 – \ln 2 \quad \cmark \end{align*}

\[\lim_{x \to 0}\dfrac{\cos(2x)-\cos(x)}{x} \to \dfrac{1-1}{0} = \dfrac{0}{0}\] Since this limit is in the form of $\dfrac{0}{0},$ we can apply L’Hôpital’s Rule: \begin{align*} \lim_{x \to 0}\dfrac{\cos(2x)-\cos(x)}{x} &= \lim_{x \to 0}\dfrac{\frac{d}{dx}[\cos(2x)-\cos(x)]}{\frac{d}{dx}(x)} \\[8px] &= \lim_{x \to 0}\dfrac{-2\sin(2x)-(-\sin(x))}{1} \\[8px] &= \dfrac{-2\cancelto{0}{\sin(0)}+\cancelto{0}{\sin(0)}}{1} \\[8px] &= \dfrac{0}{1} = 0 \quad \cmark \end{align*}

$$\lim_{x\to 0^+} x \ln x \to 0 \cdot (-\infty)$$ so we cannot use L’Hôpital’s Rule to immediately evaluate this limit.

We can, however, rewrite the first term: $x = \dfrac{1}{1/x}$. We do this because $\displaystyle{\lim_{x\to 0^+} {1/x} \to \infty}$, and so when we consider the entire expression $x \ln x =\dfrac{1}{1/x} \ln x $ we have $$\lim_{x\to 0^+} x \ln x = \lim_{x\to 0^+} \frac{\ln x}{\frac{1}{x}} \to \dfrac{-\infty}{\infty}$$ and so we can use L’Hôpital’s Rule.

\begin{align*} \lim_{x\to 0^+} x \ln x &= \lim_{x\to 0^+} \frac{\ln x}{\frac{1}{x}} \\ \\ &= \lim_{x\to 0^+} \frac{\frac{d}{dx}(\ln x)}{\frac{d}{dx}(\frac{1}{x})} \\ \\ &= \lim_{x\to 0^+} \frac{(\frac{1}{x})}{-\frac{1}{x^2}} \\ \\ &= \lim_{x\to 0^+} (-x) = 0 \quad \cmark \end{align*}

$$\lim_{x\to \infty} x^2 e^{-x} \to \infty \cdot 0$$ so we cannot use L’Hôpital’s Rule to immediately evaluate this limit.

We can, however, rewrite the second term: $e^{-x} = \dfrac{1}{e^x}$. We then have $$\lim_{x\to \infty} x^2 e^{-x} = \lim_{x\to \infty} \frac{x^2}{e^x}$$ which is in the form $\dfrac{\infty}{\infty}$ and so we can use L’Hôpital’s Rule. \begin{align*} \lim_{x\to \infty} x^2 e^{-x} &= \lim_{x\to \infty} \frac{x^2}{e^x} \\ \\ &= \lim_{x\to \infty} \frac{\frac{d}{dx}(x^2)}{\frac{d}{dx}(e^x)} \\ \\ &= \lim_{x\to \infty} \frac{2x}{e^x} \end{align*} This is again in the form $\dfrac{\infty}{\infty}$, so let’s again use L’Hôpital’s Rule:
\begin{align*} \phantom{\lim_{x\to \infty} x^2 e^{-x}} &= \lim_{x\to \infty} \frac{\frac{d}{dx}(2x)}{\frac{d}{dx}(e^x)} \\ \\ &= \lim_{x\to \infty} \frac{2}{e^x} \\ \\ &= 0 \quad \cmark \end{align*}

$$\lim_{x\to 0}\left(\frac{1}{x} – \frac{1}{\sin x} \right) \to \infty – \infty$$ so we cannot immediately use L’Hôpital’s Rule.

We can, however, put everything over a common denominator, and then examine that limit: \begin{align*} & \hphantom{= \lim_{x\to 0}\left(\frac{-\sin x}{\cos x + (\cos x – x \sin x)} \right)} \\ \lim_{x\to 0}\left(\frac{1}{x} – \frac{1}{\sin x} \right) &= \lim_{x\to 0}\left(\frac{\sin x – x}{x \sin x} \right) \\ \\ \end{align*} This limit is in the form $\dfrac{0}{0}$, and so we can apply L’Hôpital’s Rule: \begin{align*} \hphantom{\lim_{x\to 0}\left(\frac{1}{x} – \frac{1}{\sin x} \right)} & \hphantom{= \lim_{x\to 0}\left(\frac{-\sin x}{\cos x + (\cos x – x \sin x)} \right)} \\ &= \lim_{x\to 0}\left(\frac{\frac{d}{dx}(\sin x – x)}{\frac{d}{dx}(x \sin x)} \right) \\ \\ &= \lim_{x\to 0}\left(\frac{\cos x – 1}{\sin x + x \cos x} \right) \end{align*} This limit is again in the form $\dfrac{0}{0}$, so let’s applyL’Hôpital’s Rule again:
\begin{align*} \phantom{\lim_{x\to 0}\left(\frac{1}{x} – \frac{1}{\sin x} \right)} &= \lim_{x\to 0}\left(\frac{\frac{d}{dx}(\cos x – 1)}{\frac{d}{dx}(\sin x + x \cos x)}\right) \\ \\ &= \lim_{x\to 0}\left(\frac{-\sin x}{\cos x + (\cos x – x \sin x)} \right) \\ \\ &= \frac{0}{1 + (1 -0)} = 0 \quad \cmark \end{align*}

$$\lim_{x\to 0}(\cot x – \csc x) = \lim_{x\to 0}(\frac{\cos x}{\sin x} – \frac{1}{\sin x}) \to \infty – \infty$$ so we cannot immediately use L’Hôpital’s Rule.

We can, however, put everything over a common denominator, and then examine that limit: \begin{align*} \lim_{x\to 0}(\cot x – \csc x) &= \lim_{x\to 0}\left(\frac{\cos x}{\sin x} – \frac{1}{\sin x}\right) \\ \\ &= \lim_{x\to 0}\left(\frac{\cos x -1}{\sin x} \right) \end{align*} This limit is in the form $\dfrac{0}{0}$, and so we can apply L’Hôpital’s Rule: \begin{align*} \phantom{\lim_{x\to 0}(\cot x – \csc x) } &= \lim_{x\to 0}\left(\frac{\frac{d}{dx}(\cos x -1)}{\frac{d}{dx}(\sin x)}\right) \\ \\ &= \lim_{x\to 0}\left(\frac{-\sin x }{\cos x}\right) \\ \\ &= \frac{0}{1} = 0 \quad \cmark \end{align*}

$$\lim_{x\to \infty}\left(\sqrt{x^2 +x} -x \right) \to \infty – \infty$$ so we cannot use L’Hôpital’s Rule.

Let’s proceed as we would have when we first studied limits, and rationalize the expression: \begin{align*} \lim_{x\to \infty}\left(\sqrt{x^2 + x} -x \right) &= \lim_{x\to \infty}\left(\sqrt{x^2 + x} -x \right)\cdot \frac{\sqrt{x^2 + x} + x}{\sqrt{x^2 + x} + x} \\ \\ &= \lim_{x\to \infty}\frac{(x^2 +x) -x^2}{ \sqrt{x^2 +x} + x} \\ \\ &= \lim_{x\to \infty}\frac{x}{\sqrt{x^2 +x} + x} \end{align*} While this limit is now in the form $\frac{\infty}{\infty}$ and so we could apply L’Hôpital’s Rule, doing so turns rather messy. It’s thus easier to continue to proceed as we did when first studying limits, and divide each term by the largest power that appears in the denominator, x: \begin{align*} \hphantom{\lim_{x\to \infty}\left(\sqrt{x^2 + x} -x \right)} & \hphantom{ \lim_{x\to \infty}\left(\sqrt{x^2 + x} -x \right)\cdot \frac{\sqrt{x^2 + x} + x}{\sqrt{x^2 + x} + x}} \\ &= \lim_{x\to \infty}\frac{\dfrac{x}{x}}{\sqrt{\dfrac{x^2 -x}{x^2}} + \dfrac{x}{x}} \\ \\ &= \lim_{x\to \infty}\frac{1}{\sqrt{1 – \frac{1}{x} } + 1} \\ \\ &= \frac{1}{\sqrt{1 + 0} + 2} = \frac{1}{2} \quad \cmark \end{align*}

Note that we actually didn’t use L’Hôpital’s Rule in this problem. Just because we’ve introduced a new tool, don’t forget the old approaches. Sometimes they’re easier to use.

$$\lim_{x \to \infty}x^{1/x} \to \infty^0$$ so we cannot use L’Hôpital’s Rule immediately. Let’s follow the steps outlined above:

1. $y = x^{1/x}$.
2. $\ln y = \ln x^{1/x} = \dfrac{1}{x} \ln x$
3. \begin{align*} \lim_{x\to \infty}\ln y &= \lim_{x \to \infty}\frac{\ln x}{x} \end{align*} This limit is in the form $\frac{\infty}{\infty}$, and so we can apply L’Hôpital’s Rule:
\begin{align*} \phantom{\lim_{x\to \infty}\ln y } &=\lim_{x \to \infty}\frac{ \frac{d}{dx}(\ln x)}{\frac{d}{dx}(x)} \\ \\ &=\lim_{x \to \infty}\frac{\frac{1}{x}}{1} = 0 \\ \\ \lim_{x\to \infty}\ln y &= 0 \end{align*}
4. Recall that $y = e^{\ln y}$. Then \begin{align*} \lim_{x \to \infty}y &= \lim_{x \to \infty}e^{\ln y} &&\text{[We found in Step 3 that $\lim_{x\to \infty}\ln y = 0$]} \\ \\ &= e^0 = 1 \\ \\ \lim_{x\to \infty}x^{1/x} &= 1 \quad \cmark \end{align*}

$$\lim_{x\to 0^+}x^x \to 0^0$$ so we cannot use L’Hôpital’s Rule immediately. Let’s follow the steps outlined above:

1. $y = x^x$.
2. $\ln y = \ln x^x = x \ln x$
3. We found $\displaystyle{ \lim_{x\to 0^+} x \ln x}$ in a problem above. . . but for completeness will find it again here. The limit is in the form $0 \cdot -\infty$, and we can rewrite to be in the form $-\dfrac{\infty}{\infty}$ by noting that $x = \dfrac{1}{1/x}$: \begin{align*} \lim_{x\to 0^+} \ln y &= \lim_{x\to 0^+} x \ln x \\ \\ &= \lim_{x\to 0^+} \frac{\ln x}{\frac{1}{x}} \end{align*} This limit is in the form $-\dfrac{\infty}{\infty}$, and so we can use L’Hôpital’s Rule: \begin{align*} \phantom{\lim_{x\to 0^+} \ln y} &= \lim_{x\to 0^+} \frac{\frac{d}{dx}(\ln x)}{\frac{d}{dx}(\frac{1}{x})} \\ \\ &= \lim_{x\to 0^+} \frac{(\frac{1}{x})}{-\frac{1}{x^2}} \\ \\ &= \lim_{x\to 0^+} (-x) = 0 \\ \\ \lim_{x\to 0^+} \ln y &= 0 \end{align*} 4. Recall that $y = e^{\ln y}$. Then \begin{align*} \lim_{x\to 0^+}y &= \lim_{x\to 0^+}e^{\ln y} &&\text{[We found in Step 3 that $\lim_{x\to 0^+}\ln y = 0$]} \\ \\ &= e^0 = 1 \\ \\ \lim_{x\to 0^+}x^x &= 1 \quad \cmark \end{align*}

$$\lim_{x\to 0^+}(\sin x)^{\tan x} \to 0^0$$ so we cannot use L’Hôpital’s Rule immediately. Let’s follow the steps outlined above:

1. $y = (\sin x)^{\tan x}$.
2. $\ln y = \ln (\sin x)^{\tan x} = \tan x \cdot \ln (\sin x)$
3. \begin{align*} \lim_{x\to 0^+} \ln y &= \lim_{x\to 0^+} \left( \tan x \cdot \ln (\sin x) \right) \end{align*} This limit is in the form $0 \cdot (-\infty)$, but we can rewrite it to put it in the form $-\dfrac{\infty}{\infty}$ by noting that $\tan x = \dfrac{1}{\cot x}$: \begin{align*} \phantom{\lim_{x\to 0^+} \ln y} &= \lim_{x\to 0^+} \frac{\ln (\sin x) }{\cot x} \\ \\ \end{align*} This limit is in the form $-\dfrac{\infty}{\infty}$, and so we can use L’Hôpital’s Rule: \begin{align*} \phantom{ \lim_{x\to 0^+} \ln y} &= \lim_{x\to 0^+} \frac{\frac{d}{dx}(\ln (\sin x) )}{\frac{d}{dx}(\cot x)} \\ \\ &= \lim_{x\to 0^+} \frac{\left(\frac{1}{\sin x} \cdot \cos x \right)}{-\csc^2 x} \\ \\ &= \lim_{x\to 0^+}\left( \frac{1}{\sin x} \cdot \cos x \right)(-\sin^2 x) \\ \\ &= \lim_{x\to 0^+}\left(-\cos x \, \sin x \right)\\ \\ &= -1 \cdot 0 \\ \\ \lim_{x\to 0^+} \ln y &= 0 \end{align*} 4. Recall that $y = e^{\ln y}$. Then \begin{align*} \lim_{x\to 0^+}y &= \lim_{x\to 0^+}e^{\ln y} &&\text{[We found in Step 3 that $\lim_{x\to 0^+}\ln y = 0$]} \\ \\ &= e^0 = 1 \\ \\ \lim_{x\to 0^+}(\sin x)^{\tan x} &= 1 \quad \cmark \end{align*}

$$\lim_{x\to \infty}\left(1 + \frac{1}{x} \right)^x \to 1^\infty$$ so we cannot use L’Hôpital’s Rule immediately. Let’s follow the steps outlined above:

1. $y = \left(1 + \frac{1}{x} \right)^x$
2. $\ln y = \ln \left(1 + \frac{1}{x} \right)^x = x \ln \left(1 + \frac{1}{x} \right)$
3. \begin{align*} \lim_{x\to \infty}\ln y &= \lim_{x\to \infty}\left[x \ln \left(1 + \frac{1}{x} \right) \right] \end{align*} This limit is in the form $\infty \cdot 0$, but we can rewrite it to put it in the form $\frac{0}{0}$ by noting that $x = \frac{1}{1/x}$: \begin{align*} \phantom{ \lim_{x\to \infty}\ln y} &= \lim_{x\to \infty}\frac{ \ln \left(1 + \frac{1}{x} \right)}{\frac{1}{x}} \\ \\ \end{align*} This limit is in the form $\frac{0}{0}$, and so we can apply L’Hôpital’s Rule: \begin{align*} \phantom{ \lim_{x\to \infty}\ln y} &= \lim_{x\to \infty}\frac{ \frac{d}{dx}\left( \ln \left(1 + \frac{1}{x} \right) \right)}{\frac{d}{dx} \left( \frac{1}{x}\right)} \\ \\ &= \lim_{x\to \infty}\frac{\dfrac{1}{\left(1 + \frac{1}{x}\right) }\cdot \left(-\dfrac{1}{x^2}\right)}{\left(-\dfrac{1}{x^2}\right)} \\ \\ &= \lim_{x\to \infty}\dfrac{1}{\left(1 + \frac{1}{x}\right) } \\ \\ \lim_{x\to \infty}\ln y &= 1 \end{align*} 4. Recall that $y = e^{\ln y}$. Then \begin{align*} \lim_{x\to \infty}y &= \lim_{x\to \infty}e^{\ln y} &&\text{[We found in Step 3 that $\lim_{x\to \infty}\ln y = 1$]} \\ \\ &= e^1 = e \\ \\ \lim_{x\to \infty} \left(1 + \frac{1}{x} \right)^x &= e \quad \cmark \end{align*}