Unable to calculate limit without getting an indeterminate form

Question

I initially need to solve this: $$\lim_{x\to 0^+} \frac{e^{-1/x}}{x^k},\; \text{ where }k\in\mathbb{N}\cup\{0\}.$$

When I substitute $y=\dfrac1x$ then I get: $\lim\limits_{y\to\infty} \dfrac{y^k}{e^y} $

I'm unable to calculate this limit. Whatever I do, I get an indeterminate limit, even with L'Hospital's rule

Answer 1

For $k\neq0$ we have $$\displaystyle\lim_{x\to0^+}\frac{e^{-1/x}}{x^k}$$ $$=\displaystyle\lim_{x\to0^+}\frac{1}{e^{1/x}x^k}$$ $$=\left(\displaystyle\lim_{x\to0^+}\frac{1}{e^{1/x}}\right)\left(\displaystyle\lim_{x\to0^+}\frac{1}{x^k}\right)$$ $$=\left(\frac{1}{e^{\infty}}\right)\left({\frac{1}{\infty^k}}\right)$$ $$=0\times0$$ $$=0$$

In the case where $k=0$ we'd get the same answer since the expression becomes $$\displaystyle\lim_{x\to0^+}{e^{-1/x}}=e^{-\infty}=0.$$

Therefore $\displaystyle\lim_{x\to0^+}\frac{e^{-1/x}}{x^k}=0$ for all $k\in\mathbb{N}\cup\{0\}$.

Answer 2

Let compute the limit of its logarithm.

$$\lim_{y\to+\infty}(k\ln(y)-y)$$

which becomes

$$\lim_{y\to+\infty}y(k\frac{\ln(y)}{y}-1)=-\infty$$

so, your limit is zero.