How Does Differentiating Affect the Limit of an Integral Expression?

[transgalactic]

<br /> \lim_{x-&gt;+\infty} \frac{(\int_{0}^{x}e^{t^2}dt)^2}{e^{2x^2}}<br />
i was told to differentiate the integral in order to cancel it
but i don't have 0/0 infinity/infinity form
in order to differentiate the numerator and denominator.

 

[phreak]

Of course you have infinity/infinity form. Obviously, \lim_{x\to +\infty} e^{2x^2} = +\infty and \lim_{x\to +\infty} \left( \int^x_0 e^{t^2} dt \right)^2 is infinity as well, since e^{x^2} diverges as x approaches infinity.

 

[transgalactic]

<br /> \lim_{x\to +\infty} \left( \int^x_0 e^{t^2} dt \right)^2<br />
so you are saying that when we do an ante derivative to the integral we input infinity there
so it goes to infinity

<br /> \lim_{x-&gt;+\infty} \frac{(\int_{0}^{x}e^{t^2}dt)^2}{e^{2x^2}}=\lim_{x-&gt;+\infty} \frac{(e^{x^2})^2}{4xe^{2x^2}}=\lim_{x-&gt;+\infty} \frac{1}{4x}=0<br />
is it correct??

 

[HallsofIvy]

<br /> \lim_{x\to +\infty} \left( \int^x_0 e^{t^2} dt \right)^2<br />
so you are saying that when we do an ante derivative to the integral we input infinity there
so it goes to infinity

"Anti-derivative to the integral"? An integral IS an anti-derivative. It should be clear that e^{t^2} is greater than 1 for all t> 1 so the integral must be unbounded.

<br /> \lim_{x-&gt;+\infty} \frac{(\int_{0}^{x}e^{t^2}dt)^2}{e^{2x^2}}=\lim_{x-&gt;+\infty} \frac{(e^{x^2})^2}{4xe^{2x^2}}=\lim_{x-&gt;+\infty} \frac{1}{4x}=0<br />
is it correct??

No, you've differentiated wrong. The derivative of (f(x))² is 2 f(x) f'(x), not (f'(x))².