What is $ \lim\limits_{x\to\infty}(\int_0^x e^{t^2}dt)^{\frac{1}{x^2}} $ ？

Question

$ \lim\limits_{x\to\infty}(\int_0^x e^{t^2}dt)^{\frac{1}{x^2}} \\$

my idea :

$$e^{\lim\limits_{x\to\infty}\ln(\int_0^x e^{t^2}dt)^{\tfrac{1}{x^2}}}=e^{\lim\limits_{x\to\infty}\frac{\ln(\int_0^x e^{t^2} d t)}{x^2}} $$

$$ \lim\limits_{x\to\infty}\frac{\ln(\int_0^x e^{t^2} d t)}{x^2} =\lim\limits_{x\to\infty}\frac{\frac{e^{x^2}}{\int_0^x e^{t^2} d t}} {2x} = \frac{2}{2}=1$$

I don't quite understand some deformations and properties of variable upper bound integrals, thanks for your help!

Answer 1

Let $$L=\lim_{x\to\infty}\left(\int_0^{x}e^{t^2}\mathrm{d}t\right)^{\frac{1}{x^2}}.$$ Then, $$\ln(L)=\lim_{x\to\infty}\frac{\ln\left(\int_0^{x}e^{t^2}\mathrm{d}t\right)}{x^2}.$$ Apply the LH rule, the chain rule, and the fundamental theorem of calculus to get: $$\ln(L)=\lim_{x\to\infty}\frac{e^{x^2}}{2x \int_0^{x}e^{t^2}\mathrm{d}t}.$$ Apply the LH rule and the fundamental theorem of calculus again: $$\ln(L)=\lim_{x\to\infty}\frac{2xe^{x^2}}{2xe^{x^2}+2\int_0^{x}e^{t^2}\mathrm{d}t}=\lim_{x\to\infty}1-\frac{\int_0^{x}e^{t^2}\mathrm{d}t}{xe^{x^2}+\int_0^{x}e^{t^2}\mathrm{d}t}.$$ Apply the LH rule and the fundamental theorem of calculus one last time: $$\ln(L)=1-\lim_{x\to\infty}\frac{e^{x^2}}{2e^{x^2}(x^2+1)}=1-\lim_{x\to\infty}\frac{1}{2x^2+2}=1-0=1.$$ So, if $\ln(L)=1,$ we have that $L=e.$

Answer 2

From $$e^{\lim\limits_{x\to\infty}\frac{\ln(\int_0^x e^{t^2} d t)}{x^2}}, $$ applying L'Hopital $$\frac{e^{x^2}}{(\int_0^x e^{t^2} d t)2x}=\frac{2xe^{x^2}}{2xe^{x^2}+2\int_0^x e^{t^2}dt}=\frac{2e^{x^2}+4x^2e^{x^2}}{2e^{x^2}+4x^2e^{x^2}+4e^{x^2}}=\frac{4x^2+2}{4x^2+6}\to 1,$$ thus the limit is $e$.

Answer 3

A slightly different approach from L'Hospital:

You can see that for large $x$, over a significant (i.e., $\gg 1$ length) region of the interval $[0,x],$ the integrand is almost as large as $e^{x^2}$. Of course, the integrand is never larger than this. This suggests that the limit is around $(e^{x^2})^{1/x^2} = e.$

More concretely, we trivially have $\int_0^x e^{t^2} \le x e^{x^2}$. Further, note that for $t > x - \sqrt{x},$ we have $ t^2 = x^2 -2x^{3/2} + x \ge x^2 - 2 x^{3/2}$. Therefore, $$ \int_0^x e^{t^2} \ge \int_{x - \sqrt{x}}^x e^{t^2} \ge e^{x^2 - 2x^{3/2}} \cdot \sqrt{x}$$

So, we have the intequalities $$ x^{1/2x^2} e^{-2/\sqrt{x}} \cdot e \le \left( \int_0^x e^{t^2} \right)^{1/x^2} \le x^{1/x^2} \cdot e.$$

Now, as $x \to \infty, x^{1/x^2} \to 1,$ and $e^{-1/\sqrt{x}} \to 1.$ So by squeezing, we conclude that the limit in question is $e$.