Evaluating $\lim_{x\to\infty}\frac{\int_0^{2x}\sqrt{1+t^2}dt}{x^2}$

Question

How do I evaluate the following limit? $$\lim_{x\to\infty}\frac{\int_0^{2x}\sqrt{1+t^2}dt}{x^2}$$

What I've noticed so far is that due to the limit going to infinity, the integral is indefinite. Since when you work it out, you'd get an indeterminite form, I've tried L'Hopital rule but that didn't work either.

Can someone give me a tip?

Answer 1

HINT

In that case we can apply l'Hopital to obtain

$$\lim_{x\to\infty}\frac{\int_0^{2x}\sqrt{1+t^2}dt}{x^2}=\lim_{x\to\infty}\frac{2\sqrt{1+4x^2}}{2x}$$

indeed recall that

$$f(x)=\int_{a(x)}^{b(x)}g(t) dt\implies f'(x)=g(b(x))\cdot b'(x)-g(a(x))\cdot a'(x)$$

Answer 2

Note that $$\int_{0}^{2x}\sqrt{1+t^2}\,dt \le \int_{0}^{2x}\sqrt{1+t^2+2t}\,dt = \int_{0}^{2x}t+1\,dt=2x^2+2x.$$ Also, $$\int_{0}^{2x}\sqrt{1+t^2}\,dt \ge \int_{0}^{2x}\sqrt{1+t^2-2t}\,dt \ge \int_{0}^{2x}t-1\,dt=2x^2-2x.$$ Hence, by squeeze theorem, we have $$\lim_{x \to \infty}\frac{\int_{0}^{2x}\sqrt{1+t^2}\,dt}{x^2} = 2.$$