Evaluating $ \lim_{x \to 0^+} (\ln (x)+e^{\frac{-1}{x}})/( \frac{1}{x^2} + x\sin \left(\frac{-1}{x^2} \right))$

Question

Any ideas? $$ \lim_{x \to 0^+} \frac{\ln (x)+e^{\frac{-1}{x}}}{ \frac{1}{x^2} + x\sin \left(\frac{-1}{x^2} \right)}$$ L'Hopital didn't really help, maybe taylor?

Thanks

Answer 1

An idea

$$0\xleftarrow[0^+\leftarrow x]{}\frac{\log x}{\frac1{x^2}+1}\le\frac{\log x+e^{-1/x}}{\frac1{x^2}+x\sin\left(-\frac1{x^2}\right)}\le\frac{\log x+1}{\frac1{x^2}-1}\xrightarrow[x\to0^+]{}0$$

The right-hand and left-hand limits follow from applying l'Hospital, say:

$$(*)\stackrel{l'H}=\lim_{x\to0^+}\frac{\frac1x}{-\frac2{x^3}}=-\lim_{x\to0^+}\frac{x^2}2=0$$

and the inequalities follow since for $\;x\;$ pretty close to $\;0^+\;$ we have that

$$\begin{cases}0<e^{-1/x}<1\\{}\\-1\le-x\le x\sin\left(-\frac1{x^2}\right)\le x\le 1\end{cases}$$

Answer 2

hint: the answer is $0$. Just look at the first term of both the numerator and denominator and it equals to: $\dfrac{\ln x}{\dfrac{1}{x^2}}$, and use L'hospitale here because the other terms go to $0$.

Answer 3

Using KfSsOc's excellent hint, we can see that $\lim_{x \to 0^{+}} e^{-1/x} = 0$ and $\lim_{x\to 0^{+}} x \sin \left(-1/x^2\right) = 0$ Hence for small $x$ we have $$\frac{\ln x+e^{\frac{-1}{x}}}{ \frac{1}{x^2} + x\sin \left(\frac{-1}{x^2} \right)} \approx \frac{\ln x + 0}{1/x^2 + 0}$$

Hence we need only determine $$\lim_{x \to 0^{+}} \frac{\ln x}{1/x^2} = \lim_{x\to 0^{+}} -\frac{1/x}{2/x^3} = \lim_{x\to 0^{+}} \frac{x^2}{2} = 0$$

Where we used L'Hopital once above.

Answer 4

The expression equals

$$\tag 1 \frac{x^2\ln x + x^2e^{-1/x}}{1 +x^3\sin(-1/x^2)}.$$

The denominator $\to 1,$ so we need only find the limit of the numerator. Clearly $x^2e^{-1/x}\to 0.$ That $x^2\ln x \to 0$ is a basic result that is best understood without the trickery of L'Hopital. So the limit in $(1)$ is $0.$