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 August 11th, 2019, 03:49 AM #1 Senior Member   Joined: Dec 2015 From: somewhere Posts: 592 Thanks: 87 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.
 August 11th, 2019, 05:32 AM #2 Math Team     Joined: Jul 2011 From: Texas Posts: 3,002 Thanks: 1587 L'Hopital yields $4(\ln{2} - 1)$ Thanks from topsquark and idontknow
August 11th, 2019, 10:36 AM   #3
Senior Member

Joined: Dec 2015
From: somewhere

Posts: 592
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Quote:
 Originally Posted by skeeter 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 ?

 August 11th, 2019, 12:10 PM #4 Senior Member   Joined: Jun 2019 From: USA Posts: 120 Thanks: 40 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.
 August 11th, 2019, 10:55 PM #5 Global Moderator   Joined: Dec 2006 Posts: 20,921 Thanks: 2203 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). Thanks from greg1313, topsquark and idontknow

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