How can I calculate this limit? ( I tried l'Hopital and failed )

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

• How did l'hopsital fail? – qwr May 28 '14 at 23:26
• If you apply it twice then it almost returns to it's original form. – Denis May 28 '14 at 23:37
• 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.) – Martin Sleziak Jun 9 '14 at 12:00

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.

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 . . .

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

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.

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