Limit of $\frac{1}{x}\int_{x}^{2x}e^{-t^2}dt$ when $x\to0$

Question

I am trying to find the following limit: $$\lim_{x\to 0} \frac{1}{x}\int_{x}^{2x}e^{-t^2}dt.$$

I tried to solve by using the Gaussian function, but in Gaussian function the usual limits of integration are from $-\infty$ to $\infty$ and with a certain upper bound $a$ the result of a Gaussian function ended up involving the erf function. I don't know how to find this limit or even if it exists or not by using these.

Answer 1

There is no need of De l'Hopital theorem: in a neighbourhood of zero, $e^{-t^2}=1-t^2+o(t^3)$, hence:

$$ \frac{1}{x}\int_{x}^{2x}e^{-t^2}\,dt = \frac{1}{x}\left(x-\frac{7}{3}x^3+o(x^4)\right) = 1 +O(x^2) $$ and the limit of the RHS as $x\to 0$ is clearly $\color{red}{1}$.

Answer 2

Let $g(x)$ be a primitive of $e^{-x^2}$. Then

$$\lim_{x\to 0}\frac{g(2x)-g(x)}x=\lim_{x\to 0}\left(2\frac{g(2x)-g(0)}{2x}-\frac{g(x)-g(0)}x\right)=2g'(0)-g'(0)=1.$$

Answer 3

$$\frac{\int_{x}^{2x}e^{-t^2}dt}{x}=\frac{\int_{0}^{2x}e^{-t^2}dt+\int_{x}^{0}e^{-t^2}dt}{x}=\frac{\int_{0}^{2x}e^{-t^2}dt-\int_{0}^{x}e^{-t^2}dt}{x}$$

Now use L'Hospitals Rule and the fundamental Theorem of Calculus.

$$\lim_{x\to 0}\frac{\int_{x}^{2x}e^{-t^2}dt}{x}=\lim_{x\to 0}\frac{\int_{0}^{2x}e^{-t^2}dt-\int_{0}^{x}e^{-t^2}dt}{x}$$ $$=\lim_{x\to 0}\frac{2e^{-(2x)^2}-e^{-x^2}}{1}=1$$

Answer 4

Using the integral Mean Value Theorem, $$ \frac{1}{x}∫_x^{2x} f = \frac{∫_0^{2x} f - ∫_0^xf}{2x-x} = f(\xi(x))$$ for some $\xi(x) ∈ [x,2x]$. We then have $\xi(x) \xrightarrow[x→ 0]{} 0$, so by continuity of $f$, the result follows.

Answer 5

WLOG, we assume that $x>0$. Then \begin{align*} \frac{1}{x}\int_x^{2x} e^{-t^2} dt &= \int_1^2 e^{-(ux)^2} du. \end{align*} Then \begin{align*} \lim_{x\rightarrow 0}\frac{1}{x}\int_x^{2x} e^{-t^2} dt &= \lim_{x\rightarrow 0}\int_1^2 e^{-(ux)^2} du\\ &= \int_1^2 \lim_{x\rightarrow 0} e^{-(ux)^2} du\\ &= 1. \end{align*}