Rodrigo L.

asked • 08/28/20

What is the limit of the following problem, using L"Hopital's Rule

I am trying to find the limit of the following problem using L'Hopital's Rule:

limx->(infinite) x2-sqrt(x4+16x2)

Harsh P.

Hi Rodrigo, L'Hopital's rule is typically used when the numerator and denominator of a limit come out to be indeterminate. Is there a typo in your question? I would like to be able to help you out!
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08/28/20

Tom K.

Yefim's work is flawless, as always. Note that we can get an idea of the answer by completing the square on the second expression. We have sqrt(x^4 + 16x^2 + 64 -64), which is approximately x^2 + 8 - we can show that it would be x^2 + 8 - 64/(2x^2) + ... , which quickly goes to x^2 + 8. If you wonder why -64/(2x^2), recall that the derivative of sqrt(x) is 1/(2 sqrt(x)).
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08/28/20

Kevin S.

tutor
It was so long I wonder if it held attention interest, but I realized I made a slight mistake before, and I don't want to have something up that's wrong. Using L'Hopital doesn't automatically mean the limit should exist. (A silly example is [ ((1/x)sin(x))/(1/x) ] ). I fixed the end of it. But it's a little dense.
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08/30/20

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