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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?

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  • $\begingroup$ Welcome to MSE! Could you please type out your limit in MathJax (tutorial is here: math.meta.stackexchange.com/questions/5020/…) so it's easier for everyone to see? MathJax is really easy to get the hang of, trust me. $\endgroup$ – Robert Howard Aug 9 '18 at 21:00
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    $\begingroup$ thanks for editing & @Robert Howard for the link, its bookmarked :) $\endgroup$ – Victor TG Aug 9 '18 at 21:04
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    $\begingroup$ You're certainly welcome; I've lost track of how many times I've returned to that page! $\endgroup$ – Robert Howard Aug 9 '18 at 21:13
  • $\begingroup$ The integral can be evaluated in closed form with$$\int \sqrt{1+t^2}\,dx=\frac12 t\sqrt{t^2+1}+\frac12 \text{arsinh}(t)+C$$ Now, evaluate this between $0$ and $2x$, divide by $x^2$, and let $x\to \infty$. $\endgroup$ – Mark Viola Aug 9 '18 at 21:49
  • $\begingroup$ Alternatively, let $L>0$ be fixed. Then,$$\frac1{x^2}\int_0^{2x} \sqrt{1+t^2}\,dt=\underbrace{\frac1{x^2}\int_0^L \sqrt{1+t^2}\,dt}_{\to 0}+\underbrace{\frac1{x^2}\int_L^{2x} \left(t+O\left(\frac1t\right)\right)\,dt}_{\to 2}$$ $\endgroup$ – Mark Viola Aug 9 '18 at 21:57
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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

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

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    $\begingroup$ The numerator should be multiplied by $2$. $\endgroup$ – Bernard Aug 9 '18 at 21:06
  • $\begingroup$ @Bernard why? I don't understand that $\endgroup$ – Victor TG Aug 9 '18 at 21:08
  • $\begingroup$ @Bernard Yes of course I lost that factor. Thanks $\endgroup$ – user Aug 9 '18 at 21:08
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    $\begingroup$ @VictorTG: One has to apply the chain rule, because the integral is from $0$ to $\color{red}2x$. $\endgroup$ – Bernard Aug 9 '18 at 21:13
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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 $$\lim_{x \to \infty}\frac{\int_{0}^{2x}\sqrt{1+t^2}\,dt}{x^2} = 2.$$

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  • $\begingroup$ Never heard about the squeeze theorem, deff going to look it up aswell! thanks :) $\endgroup$ – Victor TG Aug 9 '18 at 21:15
  • $\begingroup$ I knew that we can solve this question using Leibnitz theorem but never thought of your way out of the box $\endgroup$ – Deepesh Meena Aug 9 '18 at 21:26
  • $\begingroup$ @VictorTG You might know it by another name. Most British mathematicians call it the sandwich theorem. I'm an exception because I don't want to confuse it with an obscure result, the ham sandwich theorem. $\endgroup$ – J.G. Aug 9 '18 at 21:27
  • $\begingroup$ Very nice use of Sandwich/squeeze +1 $\endgroup$ – Paramanand Singh Aug 10 '18 at 5:10

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