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August 11th, 2019, 03:49 AM   #1
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Simple Limit

Evaluate $\displaystyle \lim_{x\rightarrow 2 } \frac{2^x - x^2 }{2x-\sqrt{8x}}$.

Last edited by greg1313; August 11th, 2019 at 06:41 PM.
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August 11th, 2019, 05:32 AM   #2
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L'Hopital yields $4(\ln{2} - 1)$
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August 11th, 2019, 10:36 AM   #3
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Quote:
Originally Posted by skeeter View Post
L'Hopital yields $4(\ln{2} - 1)$
By all online limit calculators , the limit is 0 .
How to solve the limit without l’hopital’s rule ?
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August 11th, 2019, 12:10 PM   #4
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Wolfram Alpha sometimes gives me 0 and sometimes the correct answer, depending on how exactly I word the question. Plotting the function, however, it looks smooth and continuous in the vicinity of x = 2.

I've tried factoring out the radicals a couple different ways, but so far I haven't got rid of the 0/0. L'Hôpital seems to be the way to go on this one.
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August 11th, 2019, 10:55 PM   #5
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Replacing $x$ with $2 + x$ (for convenience), the limit becomes $\displaystyle \lim_{x\to 0} \frac{4\left(e^{\ln(2)x}\right) - (2 + x)^2 }{2(x + 2)-4\sqrt{1+ x/2}}$.

Now replacing the numerator with the first non-zero term of its Maclaurin series and replacing the denominator with the first non-zero term of its Maclaurin series gives $\displaystyle \lim_{x\to 0} \frac{4x(\ln(2) - 1)}{x}$,
which is 4(ln(2) - 1).
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