Limit of $ \frac1x \int_x ^{2x}e^{-t^2}dt$

Question

What is the limit of the function

$$\lim_{x\to 0} \ \frac1x \int_x ^{2x}e^{-t^2}dt$$ ?

I tried this problem by using gamma function. I couldn't find the integral.

Answer 1

Let $g(x)=\int_x^{2x}e^{-t^2}\mathrm d t$. Your limit is $$...=\lim_{x\to 0}\frac{g(x)-g(0)}{x-0}=:g'(0).$$

Answer 2

Using L'Hospital's Rule, we have

$$\lim_{x\to 0} \frac1x \int_x ^{2x}e^{-t^2}dt=\lim_{x\to 0}(2e^{-4x^2}-e^{-x^2})=1$$

Answer 3

By the Mean Value Theorem, $\frac1{2x-x}\int_x^{2x}e^{-t^2}\,\mathrm dt = e^{-\xi^2}$ for some $\xi$ between $x$ and $2x$. We have $e^{-\xi^2}\to 1$ as $x\to 0$.

Answer 4

Observe that

$$x\cdot e^{-4x^2} \le \int_x^{2x} e^{-t^2}\,dt \le x \cdot e^{-x^2}.$$

Dividing by $x$ and using the squeeze theorem gives the limit of $1.$