Limit of a function defined as an integral

Question

What is $\lim_{x\rightarrow 0}\dfrac{1}{x}\int_x^{2x}e^{-t^2}dt$ ? I do not get any idea how to proceed?

Answer 1

By l'Hospital

$$\lim_{x\rightarrow 0} \dfrac{\int_x^{2x}e^{-t^2}dt}{x}=\lim_{x\rightarrow 0}\, (2e^{-(2x)^2}-e^{-x^2})$$

Answer 2

HINTS:

For $t\in [x,2x]$ we have

$$e^{-t^2}\le e^{-x^2}$$

and $$\left|\int_a^{b}f(t)\,dt\right|\le \max_{t\in [a,b]}|f(t)| (b-a)$$

---

Alternatively, apply L'Hospital's Rule.

Answer 3

Let $$ F(x)=\int_0^{x}e^{-t^2}dt$$ then $$ F'(x) = e^{-x^2}$$ and

$$\lim_{x\rightarrow 0}\dfrac{1}{x}\int_x^{2x}e^{-t^2}dt= \lim_{x\rightarrow 0}\dfrac{F(2x)-F(x)}{x}= \lim_{x\rightarrow 0}(2F'(2x)-F'(x))=...$$