로피탈 정리

$1$

diverges

$1/2$

$1/2$

diverges; Limit Comparison Test with $1/x$.

converges; Limit Comparison Test with $1/x^{3/2}$.

diverges; Limit Comparison Test with $1/x$.

converges; Direct Comparison Test with $xe^{-x}$.

converges; Direct Comparison Test with $e^{-x}$.

converges; Direct Comparison Test with $xe^{-x}$.

converges; Direct Comparison Test with $1/(x^2-1)$.

diverges; Direct Comparison Test with $x/(x^2+1)$.

converges; Direct Comparison Test with $1/e^x$.

converges; Limit Comparison Test with $1/e^x$.

If we plug in $x=1$ to the numerator and the denominator, we find they are both zero. So, we have an indeterminate form appropriate for L'Hopital's Rule.

$$\begin{aligned} \lim\limits_{x\rightarrow 1}\underbrace{\frac{x^3-e^{x-1}}{\sin(\pi x)}}_{\genfrac{}{}{0in}{}{\mathrm{num} \to 0}{\mathrm{den} \to 0}} =\lim\limits_{x\rightarrow 1}\frac{3x^2-e^{x-1}}{\pi\cos(\pi x)} =-\frac{2}{\pi}\end{aligned}$$

Be careful– this is not an indeterminate form!

As $x\rightarrow 0+$, the numerator $\log x\rightarrow-\infty$. That is, the numerator is becoming an increasingly huge, negative number. As $x\rightarrow 0+$, the denominator $x\rightarrow 0+$, which only serves to make the total fraction even larger, and still negative. So, $\lim\limits_{x\rightarrow 0^+}\dfrac{\log x}{x} =-\infty$.

Remark: if we had tried to use l'Hopital's Rule here, we would have come up with the wrong answer. If we differentiate the numerator and the denominator, the fraction becomes $\dfrac{\frac{1}{x}}{1}=\frac{1}{x}$, and $\displaystyle \lim_{x \to 0^+}\frac{1}{x}=\infty$. The reason we cannot apply l'Hopital's Rule is that we do not have an indeterminate form, like both numerator and denominator going to infinity, or both numerator and denominator going to zero.

We rearrange the expression to a more natural form:

$$\begin{aligned} \lim_{x\rightarrow\infty}(\log x)^2e^{-x} &=\lim_{x\rightarrow\infty}\underbrace{\dfrac{(\log x)^2}{e^{x}}}_{\genfrac{}{}{0in}{}{\mathrm{num}\to\infty}{\mathrm{den}\to\infty}} \\ \text{Both the numerator and denominator go to infinity as $x$ goes to infinity. So, we can apply l'H\^opital's Rule. In fact, we end up applying it twice.} \\ &=\lim_{x\rightarrow\infty}\underbrace{\dfrac{2\log x}{xe^{x}}}_{\genfrac{}{}{0in}{}{\mathrm{num}\to\infty}{\mathrm{den}\to\infty}} \\ &=\lim_{x\rightarrow\infty}\dfrac{2/x}{xe^{x}+e^x} \\ \text{The numerator gets smaller and smaller while the denominator gets larger and larger, so:} \\ &={0} \end{aligned}$$

$$\lim_{x\rightarrow\infty}x^2e^{-x} =\lim_{x\rightarrow\infty}\underbrace{\frac{x^2}{e^{x}}}_{\genfrac{}{}{0in}{}{\mathrm{num}\to \infty}{\mathrm{den}\to \infty}} =\lim_{x\rightarrow\infty}\underbrace{\frac{2x}{e^{x}}}_{\genfrac{}{}{0in}{}{\mathrm{num}\to\infty}{\mathrm{den}\to\infty}} =\lim_{x\rightarrow\infty}\underbrace{\frac{2}{e^{x}}}_{\genfrac{}{}{0in}{}{\mathrm{num}\to\infty}{\mathrm{den}\to\infty}} ={0}$$

$$\begin{aligned} \lim_{x\rightarrow 0}\underbrace{\frac{x-x\cos x}{x-\sin x}}_{\genfrac{}{}{0in}{}{\mathrm{num}\to0}{\mathrm{den}\to0}} &= \lim_{x\rightarrow 0}\underbrace{\frac{1-\cos x+x\sin x}{1-\cos x}}_{\genfrac{}{}{0in}{}{\mathrm{num}\to0}{\mathrm{den}\to0}} = \lim_{x\rightarrow 0}\underbrace{\frac{\sin x+\sin x+x\cos x}{\sin x}}_{\genfrac{}{}{0in}{}{\mathrm{num}\to0}{\mathrm{den}\to0}}\\ &= \lim_{x\rightarrow 0}\frac{2\cos x+\cos x-x\sin x}{\cos x} ={3} \end{aligned}$$

If we plug in $x=0$ to the numerator and denominator, both are zero, so this is a candidate for l'Hopital's Rule. However, an easier way to evaluate the limit is to factor $x^2$ from the numerator and denominator, and cancel.

$$\begin{aligned} \lim_{x\to0}\frac{\sqrt{x^6+4x^4}}{x^2\cos x} &=\lim_{x\to0}\frac{\sqrt{x^4}\sqrt{x^2+4}}{x^2\cos x}\\ &=\lim_{x\to0}\frac{x^2\sqrt{x^2+4}}{x^2\cos x}\\ &=\lim_{x\to0}\frac{\sqrt{x^2+4}}{\cos x}\\ &=\frac{\sqrt{0^2+4}}{\cos(0)}=2 \end{aligned}$$