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Is anyone able to help me with this question on l'Hopital's rule?

Use l'Hopital's rule to find the limit of the sequence $\{a_n\}_{n=1}^\infty$ with $n$-th term $\displaystyle a_n = \frac{\ln(n)}{e^n}$

Is my work correct? If not, how could I improve it?

$$\lim\limits_{n\to \infty} \frac{\ln(n)}{e^n} = \lim\limits_{n\to \infty} \frac{1/n}{e^n} = \lim\limits_{n\to \infty} \frac{1}{ne^n} = \dfrac{1}{\infty} = 0$$

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    $\begingroup$ That's perfect. You should say that you get $\infty/\infty$ from your limit, which is why you're applying L'Hospital's Rule but other than that, it's exactly correct. $\endgroup$ – Cameron Williams Mar 24 '14 at 1:05
  • $\begingroup$ My professors always told me to write that it came out as an indeterminate form, and thus we apply L'Hospital's rule. As the poster above me commented. $\endgroup$ – H5159 Mar 24 '14 at 1:15
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It is correct, but a remark is necessary.

Since the functions $x\mapsto \log x$ and $x\mapsto e^x$ are defined for $x>0$ and, with l'Hôpital's theorem, $$ \lim_{x\to\infty}\frac{\log x}{e^x}\overset{(\mathrm{H})}{=} \lim_{x\to\infty}\frac{1/x}{e^x}=\lim_{x\to\infty}\frac{1}{xe^x}=0 $$ we can conclude that, whenever we have a sequence $a_n$ such that $\lim_{n\to\infty}a_n=\infty$, $$ \lim_{n\to\infty}\frac{\log a_n}{e^{a_n}}=0. $$ The particular case follows with $a_n=n$

There are cases where a sequence $b_n$ can be written as $b_n=f(n)$, where $f$ is defined on (a subset of) the real numbers, where $$ \lim_{n\to\infty}b_n $$ exists, but $$ \lim_{x\to\infty}f(x) $$ doesn't. A simple example is with $f(x)=\sin(2\pi x)$; the sequence $b_n=f(n)$ converges, because it's constant, but $f$ doesn't have a limit at infinity.

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If $\lim\limits_{x\to \infty} f (x)=l$ and $x_n =f (n)$ then $\lim\limits_{x\to \infty}{x_n}=l$

The sequences are not differentiable.

Correct se l'Hopital's rule:

$$\lim\limits_{x\to \infty} \frac{\ln(x)}{e^x} = \lim\limits_{x\to \infty} \frac{1/x}{e^x} = \lim\limits_{x\to \infty} \frac{1}{xe^x} = \dfrac{1}{\infty} = 0$$

For $x=n$ result $$\lim\limits_{n\to \infty} \frac{\ln(n)}{e^n} =0$$

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