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 $4(\ln{2}  1)$

August 11th, 2019, 10:36 AM  #3 
Senior Member Joined: Dec 2015 From: somewhere Posts: 592 Thanks: 87  
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 nonzero term of its Maclaurin series and replacing the denominator with the first nonzero 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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