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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1Have you learnt L'hopital's rule? – Oria Gruber Jul 07 '14 at 14:43
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@OriaGruber No, I am teaching myself mathematics. Thanks, didn't know about this. – Pacha Jul 07 '14 at 14:45
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Alternately, how are you feeling about Taylor series? – Jason Knapp Jul 07 '14 at 14:45
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Sandwich also works great here. – Oria Gruber Jul 07 '14 at 14:50
5 Answers
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Try L'Hospital's Rule: $$ \lim_{x\to \infty}\frac{f(x)}{g(x)} = \lim_{x\to \infty}\frac{f'(x)}{g'(x)}. $$
draks ...
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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.
Oria Gruber
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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 $$ \begin{align} \lim_{x\to\infty}\frac1xe^x &\ge\lim_{x\to\infty}\frac1x\left(1+x/2\right)^2\\ &\ge\lim_{x\to\infty}x/4\\ &\to\infty \end{align} $$
robjohn
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Note also that by L'Hospital's Rule, $e^x$ grows quicker than any polynomial as $x$ approaches $\infty$.
rehband
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1In fact, for $x\ge0$, we can modify the idea in my answer to get $$e^x\ge\left(1+\frac xn\right)^n$$ – robjohn Jul 08 '14 at 09:52
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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.
najtofni
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1In my humble opinion, it would be more appropriate to post this as a comment. Might I suggest you read the about page – gebruiker Jul 07 '14 at 18:34