Finding $\lim_{\varepsilon\to 0}\int_{\varepsilon}^{\frac{\varepsilon}{1+\varepsilon}}\frac{e^{-x^2}}{x^2}dx$ $$\lim_{\varepsilon\to 0}\int_{\varepsilon}^{\frac{\varepsilon}{1+\varepsilon}}\frac{e^{-x^2}}{x^2}dx$$
Okay this seems weird to me-- couldn't i just write down the answer 0 because $\int_{0}^{0} f(x)dx=0$? (The answer is $-1$ btw)
There was even a hint for the question: $\displaystyle \lim_{x\to0}\frac{e^{-x^2}-1}{x^2}$ may be useful.
Am I supposed to go  $$\displaystyle \int_{\varepsilon}^{0}\frac{e^{x^2}}{x^2}dx+\int_{0}^{\frac{\varepsilon}{1+\varepsilon}}\frac{e^{x^2}}{x^2}dx\,\,\,?$$ And I couldn't even go L'Hospital with that.....
Any help would be appreciated ! Thanks!
Of course, it would be nice if the answer was simply 0!
 A: The value of this limit is $-1$. $$\int_{\epsilon} ^{\frac {\epsilon} {1+\epsilon}} \frac {e^{-x^2}} {x^{2}}dx =\int_{\epsilon} ^{\frac {\epsilon} {1+\epsilon}} \frac {e^{-x^2}-1} {x^{2}}dx +\int_{\epsilon} ^{\frac {\epsilon} {1+\epsilon}} \frac 1 {x^{2}} dx$$. You can easily evaluate the limit of  second term by evaluating the integral explicitly. For the first term observe that $\frac {e^{-x^2}-1} {x^{2}}$ is bounded function. For any bounded function $f$ we have $$\left|\int_{\epsilon} ^{\frac {\epsilon} {1+\epsilon}} f(x)dx\right| \leq M\left(\frac {\epsilon} {1+\epsilon}-\epsilon\right) \to 0$$ where $M$ is a bound for $|f|$.
For boundedness of $\frac {e^{-x^2}-1} {x^{2}}$ note that it is continuous on $(0,1]$ and it has  a finite limit, namely $0$, at $0$. (Use series expansion for the limit. You can also use L'Hopital's Rule if you prefer).
A: Consider
$I:=\displaystyle{\int_{\epsilon/(\epsilon+1)}^{e}} \dfrac{e^{-x^2}}{x^2}dx, \epsilon >0$;
$e^{-(\epsilon/(1+\epsilon))^2} \displaystyle{\int_{\epsilon/(\epsilon+1)}^{\epsilon}}\dfrac{1}{x^{2}}dx> I >$
$e^{-\epsilon^{2}}\displaystyle{\int_{\epsilon/(1+\epsilon)}^{\epsilon}}\dfrac{1}{x^2}dx;$
Calculate the above integral and take the limit $\epsilon \rightarrow 0^{+}$ (Squeeze Theorem).
Similarly for $\epsilon \rightarrow 0^{-}.$
A: Since $\varepsilon\to 0$, the domain of integration is very small ($\frac{\varepsilon}{1+\varepsilon}\sim \varepsilon-\varepsilon^2$). Using series
$$\frac{e^{-x^2}}{x^2}=\frac{1}{x^2}-1+\frac{x^2}{2}+O\left(x^4\right)$$
$$\int\frac{e^{-x^2}}{x^2}dx\sim -\frac{1}{x}-x+\frac{x^3}{6}+O\left(x^5\right)$$
Using the bounds,
$$I=\left(\frac{\varepsilon ^3}{6 (\varepsilon +1)^3}-\frac{\varepsilon }{\varepsilon
   +1}-\frac{\varepsilon +1}{\varepsilon } \right) -\left(\frac{\varepsilon ^3}{6}-\varepsilon -\frac{1}{\varepsilon } \right)$$
Using Taylor
$$I=-1+\varepsilon ^2-\varepsilon ^3+O\left(\varepsilon ^4\right)$$
A: I don't know if you like this method or not. By the MVT for definite integrals, there is $\xi\in(\varepsilon,\frac{\varepsilon}{1+\varepsilon})$ such that
$$\int_{\varepsilon}^{\frac{\varepsilon}{1+\varepsilon}}\frac{e^{-x^2}}{x^2}dx=e^{-\xi^2}\int_{\varepsilon}^{\frac{\varepsilon}{1+\varepsilon}}\frac{1}{x^2}dx=e^{-\xi^2}(-\frac{1}{x})\bigg|_{\varepsilon}^{\frac{\varepsilon}{1+\varepsilon}}=-e^{-\xi^2}$$
and hence
$$\lim_{\varepsilon\to 0}\int_{\varepsilon}^{\frac{\varepsilon}{1+\varepsilon}}\frac{e^{-x^2}}{x^2}dx=\lim_{\varepsilon\to 0}(-e^{-\xi^2})=-1.$$
