Can the limit $\lim_{x\to0}\left(\frac{1}{x^5}\int_0^xe^{-t^2}\,dt-\frac{1}{x^4}+\frac{1}{3x^2}\right)$ be calculated?

Question

$$\displaystyle\lim_{x\to0}\left(\frac{1}{x^5}\int_0^xe^{-t^2}\,dt-\frac{1}{x^4}+\frac{1}{3x^2}\right)$$

I have this limit to be calculated. Since the first term takes the form $\frac 00$, I apply the L'Hospital rule. But after that all the terms are taking the form $\frac 10$. So, according to me the limit is $ ∞$. But in my book it is given 1/10. How should I solve it?

Answer 1

One may recall that, as $t \to 0$, by the Taylor series expansion $$ e^{-t^2}=1-t^2+\frac{t^4}2+O(t^6) $$ giving, as $x \to 0$, $$ \int_0^xe^{-t^2}dt=x-\frac{x^3}3+\frac{x^5}{10}+O(x^7) $$ and, as $x \to 0$,

$$ \frac1{x^5}\int_0^xe^{-t^2}dt-\frac1{x^4}+\frac1{3x^2}=\frac1{10}+O(x^2) $$

from which one deduces the desired limit.

Answer 2

Hint: $e^{-u} = 1-u+u^2/2 + O(u^3).$

Answer 3

By bringing the fractions to the same denominator, start by writing the limit as $$\displaystyle\lim_{x\to0}\frac{3\int_0^xe^{-t^2}\,dt -3x+x^3 }{3x^5}$$

Now, since this is of the form $0/0$ by L'H and FTC you get $$\displaystyle\lim_{x\to0}\frac{3e^{-x^2}-3+x^2 }{15x^4}$$

From here it is easy.