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How do you solve this limit? I know this is probably really easy.

$$ \lim_{x \to ∞} \left(f(x) = (1 / x) * e ^ x\right) $$

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    $\begingroup$ Have you learnt L'hopital's rule? $\endgroup$ Jul 7, 2014 at 14:43
  • $\begingroup$ @OriaGruber No, I am teaching myself mathematics. Thanks, didn't know about this. $\endgroup$
    – Pacha
    Jul 7, 2014 at 14:45
  • $\begingroup$ Alternately, how are you feeling about Taylor series? $\endgroup$ Jul 7, 2014 at 14:45
  • $\begingroup$ Sandwich also works great here. $\endgroup$ Jul 7, 2014 at 14:50

5 Answers 5

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Try L'Hospital's Rule: $$ \lim_{x\to \infty}\frac{f(x)}{g(x)} = \lim_{x\to \infty}\frac{f'(x)}{g'(x)}. $$

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Another way would be the sandwich method:

Notice that $e^{0.5x}$ is monotonic ascending, and et's all agree that after a certain point $n$, $e^{0.5x} >x$ for all $x > n$ (Showing this is pretty easy and I will leave it to you).

And so, if $x>n$:

$$\frac{e^x}{e^{0.5x}} < \frac{e^x}{x} < e^x $$

When $x$ tends to infinity. the left side limit approaches infinity, the right side limit approaches to infinity, so what's bound in the middle must approach infinity as well.

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For all $x\in\mathbb{R}$, we have $e^{x/2}\ge1+x/2$. Therefore, for $x\gt0$, $e^x\ge\left(1+x/2\right)^2$, and so $$ \begin{align} \lim_{x\to\infty}\frac1xe^x &\ge\lim_{x\to\infty}\frac1x\left(1+x/2\right)^2\\ &\ge\lim_{x\to\infty}x/4\\ &\to\infty \end{align} $$

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Note also that by L'Hospital's Rule, $e^x$ grows quicker than any polynomial as $x$ approaches $\infty$.

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    $\begingroup$ In fact, for $x\ge0$, we can modify the idea in my answer to get $$e^x\ge\left(1+\frac xn\right)^n$$ $\endgroup$
    – robjohn
    Jul 8, 2014 at 9:52
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The L'Hospital rule is applicable here, however the other answers fail to stress its limitations. Be sure to read the wikipedia article.

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    $\begingroup$ In my humble opinion, it would be more appropriate to post this as a comment. Might I suggest you read the about page $\endgroup$
    – gebruiker
    Jul 7, 2014 at 18:34

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