Limit of integral question

[tex]
\lim_{x->+\infty} \frac{(\int_{0}^{x}e^{t^2}dt)^2}{e^{2x^2}}
[/tex]
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.
 
141
1
Of course you have infinity/infinity form. Obviously, [tex] \lim_{x\to +\infty} e^{2x^2} = +\infty [/tex] and [tex] \lim_{x\to +\infty} \left( \int^x_0 e^{t^2} dt \right)^2 [/tex] is infinity as well, since [tex]e^{x^2}[/tex] diverges as x approaches infinity.
 
[tex]
\lim_{x\to +\infty} \left( \int^x_0 e^{t^2} dt \right)^2
[/tex]
so you are saying that when we do an ante derivative to the integral we input infinity there
so it goes to infinity

[tex]
\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
[/tex]
is it correct??
 

HallsofIvy

Science Advisor
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[tex]
\lim_{x\to +\infty} \left( \int^x_0 e^{t^2} dt \right)^2
[/tex]
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 [itex]e^{t^2}[/itex] is greater than 1 for all t> 1 so the integral must be unbounded.

[tex]
\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
[/tex]
is it correct??
No, you've differentiated wrong. The derivative of (f(x))2 is 2 f(x) f'(x), not (f'(x))2.
 

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