Finding $\lim_{x \to 0^+} \frac{e^{-1/x}}{x} \cdot \int_x^1 \frac{e^{1/t}}{t} \mathrm{d}t$

Question

I am trying to find the following :

$$\lim_{x \to 0^+} \frac{e^{-1/x}}{x} \cdot \int_x^1 \frac{e^{1/t}}{t} \mathrm{d}t$$

I tried the following :

• $e^{1/t-1/x}/(xt) = \sum_{k = 0}^{\infty} \frac{(1/t-1/x)^k}{(xt)^k}$ yet I am enable to find an antiderivative of the function : $x \mapsto \frac{(1/t-1/x)^k}{(xt)^k}$...

Answer 1

Apply L'Hopital's Rule. $\lim_{x \to 0+} \frac {\int_x^{1}{e^{1/t} /t dt}} {xe^{1/x}}=\lim_{x \to 0+}\frac {-\frac 1 xe^{1/x}} {(1-\frac 1 x)e^{1/x}}=\lim_{x \to 0+}\frac 1 {1-x}=1$