On divergence of convergent looking integral

Question

How is $\displaystyle\int_1^\infty \frac{dx}{e^{1/x}}$ divergent? By ocular inspection it looks to me as if the lower limit doesn't cause any problems and neither does the upper limit as $\lim_{x\to\infty} e^{-1/x}=1$.

Answer 1

"as $\lim_{x\to\infty} e^{-1/x}=1$" is precisely the problem: since there exists $M>1$ such that $\forall x>M,\, e^{-1/x}>\frac12$, you have that, as soon that $t>M$, $$\int_1^t e^{-1/x}\,dx\ge \frac12(t-M)+\int_1^M e^{-1/x}\,dx=\frac{t-M}2+c$$ hence $$\liminf_{t\to\infty}\int_1^t e^{-1/x}\,dx\ge \lim_{t\to\infty}\frac{t-M}2+c=+\infty$$ hence the integral is divergent.

Answer 2

The fact that $\displaystyle \lim_{x\to\infty} \frac 1 {e^{1/x}} =1 $ implies that for big enough values of $x$ we have $\dfrac 1 {e^{1/x}} > \dfrac 1 2$. And therefore the integral is $\displaystyle \ge \int_a^\infty \frac 1 2 \, dx = +\infty$ (where $a$ is big enough so that $\dfrac 1 {e^{1/x}} > \dfrac 1 2$ when $x\ge a$).