How to use l'hopital rule?

Question

So I have this question

$$\lim_{x\rightarrow 0^+}\exp\left(-\frac{1}{x}\right)\ln\left(\frac{1}{x^2}\right)$$

I was wondering how to solve the limit tending to 0 ? I thought of using l'Hôpital rule, but this equation is not in the form of a fraction ? What could I do ? Thanks!

Answer 1

$$\lim_{x\rightarrow 0^+}\exp\left(-\frac{1}{x}\right)\ln\left(\frac{1}{x^2}\right)$$ $$\lim_{x\rightarrow 0^+}\frac{\ln\left(\frac{1}{x^2}\right)}{\exp\left(\frac{1}{x}\right)}$$ so $\infty/\infty$ $$\lim_{x\rightarrow 0^+}\frac{\frac{d}{dx}\ln\left(\frac{1}{x^2}\right)}{\frac{d}{dx}\exp\left(\frac{1}{x}\right)}$$ Can you take it from there?

Answer 2

Needless to appeal to L'Hospital's rule. A simple substitution $t=\dfrac1x$, and basic facts will do: $$\mathrm e^{-\tfrac1x}\,\ln\Bigl(\frac 1{x^2}\Bigr)=\frac{2\ln t}{\mathrm e^t}\qquad(t\to +\infty\; \text{ when }\;x\to 0_+). $$ Now, near $+\infty,\;\ln t=o(t)\;$ and $\;t=o(\mathrm e^t)$, so $\;\ln t=o(\mathrm e^t)$, and the limit is $0$.

Answer 3

You can use L'Hospital's rule here, but it's easier to substitute first. $t=\frac{1}{x}$ $$ \lim_{t\to \infty} e^{-t} \ln(t^2) = \lim_{t\to\infty} \frac{2\ln(t)}{e^t} = \lim_{t\to\infty}\frac{\frac{2}{t}}{e^{t}} = 0 $$