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I have to calculate this : $$ \lim_{x\to 0}\frac{2-x}{x^3}e^{(x-1)/x^2} $$ Can somebody help me?

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  • $\begingroup$ How did l'hopsital fail? $\endgroup$ – qwr May 28 '14 at 23:26
  • $\begingroup$ If you apply it twice then it almost returns to it's original form. $\endgroup$ – Denis May 28 '14 at 23:37
  • $\begingroup$ The tag (limit-theorems) is intended for questions about limit theorems in probability theory and not for questions about determining limits of sequences or functions, see the tag-wiki and the tag-excerpt. (The tag-excerpt is also shown when you are adding a tag to a question.) $\endgroup$ – Martin Sleziak Jun 9 '14 at 12:00
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Hint: It may be fruitful to substitute $\alpha = 1/x$, in which case you obtain the limit

$$ \lim_{ \alpha \rightarrow \infty} \left(2 - \frac{1}{\alpha} \right) \alpha^3 e^{\alpha - \alpha^2} $$

I should note that, here, I'm taking your limit to in fact be the limit as $x$ approaches $0$ from the positive direction. If you're intending for your limit to be two-sided, then you should think about why that would cause problems.

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Letting $w=1/x$, we have $$ \lim_{x\downarrow 0}\frac{2-x}{x^3}e^{(x-1)/x^2} = \lim_{w\to+\infty} \left(2 - \frac 1 w \right) w^3 e^{w^2\left(\frac 1 w - 1\right)} = \lim_{w\to+\infty} (2w^3 - w^2) e^{w-w^2} $$ $$ = \lim_{w\to+\infty} \frac{2w^3-w^2}{e^{w^2-w}}. $$ L'Hopital should handle that.

Maybe I'll post something on $x\uparrow 0$ later . . .

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$$\lim_{x\to0}(f(x)g(x)) = \lim_{x\to0}(f(x)) \cdot \lim_{x\to0}(g(x)) $$

With that being said you can let $f(x) = (2-x)/x^3 $ and $ g(x) = e^{(x-1)/x^2} $

I hope this helps.

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  • $\begingroup$ It doesn't because the first limit is infinite and the second limit is 0 and you can't multiply them. $\endgroup$ – Denis May 28 '14 at 23:29
  • $\begingroup$ Ah right I didn't even really think about the limits. $\endgroup$ – meh May 28 '14 at 23:31

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