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anonymous
 5 years ago
another question on limits
anonymous
 5 years ago
another question on limits

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{x \rightarrow \infty} (\int\limits_{0}^{x}e ^{x}dx)^{2}/\int\limits_{0}^{x}e ^{2x ^{2}}dx\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The numerator easily simplifies to \[ \left(e^x  1\right)^2 \] but the lower integral has no elementary antiderivative, so I'm not too sure.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If you could show that the limit of the denominator goes to infinity, you could apply L'Hopital's.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how about using l'hospital's rule?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think l'hospital's rule would solve it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well whats the integration of the denominator?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0There is no elementary solution to it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can just apply the fundamental theorem of calculus to find the derivative of numerator and denominator.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, but L'Hopital's only works if you know you have an indeterminate form. We don't know what the limit of the denominator is.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think by looking at a graph for e^(x^2), we can see that the area under this graph goes to infinity as x goes to infinity.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That works. Then after application of FToC and L'Hopital's this should be pretty easy to solve.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can any one of you solve it.. i have no idea whatsoever...lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, I got the answer to be 0. Is it right at first?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0We're going to apply L'Hopital's rule, but you have first to be aware of the Fundamental theorem of calculus, that states: \[{d \over dx}{\int\limits_{0}^{x}f{(t)}d t}=f(x)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Just take it for now :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[{d \over dx}({\int\limits_{0}^{x}e^t d. t })^2=2\int\limits_{0}^{x}e^t d t. e^x=2e^x(e^x1)\] I used chain rule here as well.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That's the derivative of the numerator.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Are you following so far?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The derivative of the denominator is just: \[{d \over dx}\int\limits_{0}^{x}e^{2t^2} d.t=e^{2x^2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0doesn't the anti derivative come out to be \[\left(e^x  1\right)^2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well what you wrote above is the value of the integral, but we're looking for the derivative of the square of the integral. That's equal to, by chain rule, 2*the integral*its derivative.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can you explain this a little.. {d \over dx}({\int\limits_{0}^{x}e^t d. t })^2=2\int\limits_{0}^{x}e^t d t. e^x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[{d \over dx}({\int\limits\limits_{0}^{x}e^t d. t })^2=2\int\limits\limits_{0}^{x}e^t d t. e^x\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Good. Now we have found both derivatives of the numerator and the denominator. So, we have: \[{2e^{2x}2e^x \over e^{2x^2}}={2e^{2x}(1{1 \over e^x}) \over e^{2x}(e^{2x^22x})}={2{2 \over e^x} \over e^{2x^22x}}\] The top goes to 2 as x goes to infinity, and the bottom goes to infinity as x goes to infinity. Hence the limit goes to 0.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The key to find what you asked about is applying both chain rule and FToC at the same time. So for chain rule assume the integral is u, then: d/du(u^2)=2u.u'. u is the value of the integral which is (e^x1). u' is the derivative of the integral which, by FToC, e^x. ans 2 is just 2 :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0To understand the FToC, look at the differentiation, for now, as the inverse operation of integration. So they "undo" each other. I hope that helps.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmm..got it.. but i didn't understand how you simplified the denominator here \[{2e^{2x}2e^x \over e^{2x^2}}={2e^{2x}(1{1 \over e^x}) \over e^{2x}(e^{2x^22x})}={2{2 \over e^x} \over e^{2x^22x}}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0When you take something is a common factor, you divide the rest by it. Right? When dividing you subtract powers, that's all.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I gotta go now. I may catch up with you later. Bye!!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh ya...how did i miss that! thanx a lot bro.. =)=)
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