$\lim_{x\to0^+}\frac{e^{-3/x}}{x^2}$

Question

$$\lim_{x\to0^+}\frac{e^{-3/x}}{x^2}$$ I used L'Hôpital's rule once and got up to $\frac3{2x^3e^{3/x}}$. I then rearranged the function to get $\frac3{(e^{3/x}/(1/2x^3))}$. From here on I used L'Hôpital's rule, rearranged the function for about $3$ times before I gave up. Can somebody pinpoint my mistake and show me the complete working? I apologise for the confusing formulas as I do know how to type them out in the right format yet.

Answer 1

Note that for $x=\frac1y$ with $y\to +\infty$

$$\frac{e^{-\frac3x}}{x^2}=y^2e^{-3y}=\frac{y^2}{e^{3y}}\to 0$$

indeed, without l'Hospital, we can simply observe that eventually $e^{3y}>y^3$ then

$$0<\frac{y^2}{e^{3y}}<\frac{y^2}{y^3}=\frac1y\to 0$$

therefore the limit follow by squeeze theorem.