# Limit of integral question

#### transgalactic

$$\lim_{x->+\infty} \frac{(\int_{0}^{x}e^{t^2}dt)^2}{e^{2x^2}}$$
i was told to differentiate the integral in order to cancel it
but i dont have 0/0 infinity/infinity form
in order to differentiate the numerator and denominator.

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#### 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

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

$$\lim_{x->+\infty} \frac{(\int_{0}^{x}e^{t^2}dt)^2}{e^{2x^2}}=\lim_{x->+\infty} \frac{(e^{x^2})^2}{4xe^{2x^2}}=\lim_{x->+\infty} \frac{1}{4x}=0$$
is it correct??

#### HallsofIvy

$$\lim_{x\to +\infty} \left( \int^x_0 e^{t^2} dt \right)^2$$
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.

$$\lim_{x->+\infty} \frac{(\int_{0}^{x}e^{t^2}dt)^2}{e^{2x^2}}=\lim_{x->+\infty} \frac{(e^{x^2})^2}{4xe^{2x^2}}=\lim_{x->+\infty} \frac{1}{4x}=0$$
is it correct??
No, you've differentiated wrong. The derivative of (f(x))2 is 2 f(x) f'(x), not (f'(x))2.

"Limit of integral question"

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