# Compute $\lim\limits_{x\to0}\left(\frac{1}{x^4}-\frac{\int_0^{x^2}e^{-u^2}du}{x^6}\right)$ without L'Hopital rule nor Taylor

I would like to compute the following $$\lim_{x\to0}\left(\frac{1}{x^4}-\frac{\displaystyle\int_0^{x^2}e^{-u^2}du}{x^6}\right)$$

I have done it using L'Hospital rule and Taylor and I obtained $1/3$

Is there a way I can compute this limit without appealing L'Hospital rule or Taylor? I am likely thinking about squeezing or a more elegant way.

• What's wrong in using a Taylor expansion? – egreg Jan 26 '18 at 22:14
• If you put $x^2=y$, the limit will be $\lim_{y\to0}\left(\frac{1}{y^2}-\frac{\displaystyle\int_0^{y}e^{-u^2}du}{y^3}\right)$ and using L'Hospital is simpler. – sdcvvc Jan 26 '18 at 22:38
• The function you're taking the limit of would be equal to $\frac{1}{x^6} \int_0^{x^2} (1 - e^{-u^2}) du$. Then bounding $1 - e^{-u^2}$ between $u^2$ and $u^2 - \frac{u^4}{2}$ should do the trick. – Daniel Schepler Jan 26 '18 at 22:48

The expression under limit can be written as $$\frac{1}{x^6}\int_{0}^{x^2}(1-e^{-u^2})\,du$$ We use the standard inequality $$1+t\leq e^t\leq \frac{1}{1-t},0\leq t <1$$ and take reciprocal to get $$1-t\leq e^{-t} \leq\frac{1}{1+t}$$ Putting $t=u^2$ where $0\leq u\leq x^2<1$ we get $$1-\frac{1}{1+u^2}\leq 1-e^{-u^2}\leq 1-(1-u^2)$$ ie $$\frac{u^2}{1+u^2}\leq 1-e^{-u^2}\leq u^2$$ Since $0\leq u\leq x^2$ we have $$\frac{u^2}{1+x^4}\leq 1-e^{-u^2}\leq u^2$$ and integrating with respect to $u$ on interval $[0,x^2]$ we get $$\frac{x^6}{3(1+x^4)}\leq \int_{0}^{x^2}(1-e^{-u^2})\,du\leq \frac{x^6}{3}$$ By Squeeze theorem the desired limit is $1/3$.
with $t^2=u$ the integral becomes
$$2\int_0^x e^{-t^4}tdt$$ $$=2\int_0^x (t-t^5+t^5\epsilon (t))dt$$ $$=x^2-\frac {x^6}{3}+x^6\epsilon (x)$$
the limit will be $$\frac {1}{3}$$