My Math Forum Please suggest solution for limit of (e^x+e^-x-2)/x^2 as x goes to zero
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 March 11th, 2017, 12:39 PM #1 Newbie   Joined: Mar 2017 From: San Ramon Posts: 1 Thanks: 0 Please suggest solution for limit of (e^x+e^-x-2)/x^2 as x goes to zero Hi All, Can you please suggest how can I solve this : Finding the limit of (e^x+e^-x-2)/x^2 as x goes to zero
 March 11th, 2017, 01:19 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,574 Thanks: 931 Math Focus: Elementary mathematics and beyond Use L'Hôpital's rule. Last edited by skipjack; March 11th, 2017 at 03:39 PM.
 March 11th, 2017, 01:25 PM #3 Global Moderator   Joined: May 2007 Posts: 6,306 Thanks: 525 You need to do it twice.
 March 11th, 2017, 03:13 PM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,939 Thanks: 2265 Math Focus: Mainly analysis and algebra This problem is perfect for Maclaurin series. Far easier than l'Hôpital too. Thanks from Joppy
 March 11th, 2017, 05:14 PM #5 Member   Joined: Jan 2016 From: Athens, OH Posts: 46 Thanks: 26 Archie, I understand you have a personal vendetta against l'hôpital, but for beginners, I think this is much easier to understand than a series solution. Besides, it's much easier to type. Last edited by skipjack; March 11th, 2017 at 07:28 PM.
 March 11th, 2017, 06:10 PM #6 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,574 Thanks: 931 Math Focus: Elementary mathematics and beyond The limit is one.
 March 11th, 2017, 06:17 PM #7 Member   Joined: Jan 2016 From: Athens, OH Posts: 46 Thanks: 26 In terms of age, I'm definitely not a beginning student. However, I managed to come up with the wrong answer in two different ways. So here's the correction:
March 11th, 2017, 07:18 PM   #8
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Quote:
 Originally Posted by johng40 Archie, I understand you have a personal vendetta against l'hôpital, but for beginners, I think this is much easier to understand than a series solution. Besides, it's much easier to type.
I don't have a personal vendetta against l'Hôpital, but I do think that it gives the least understanding of what is happening at a limit. Also, if one always immediately turns to l'Hôpital (as many do), you will have no tools left when l'Hôpital fails.

It might be easier to type if you pick the worst possible way to write the series! If you write it sensibly, it is very easy and extremely clear. Unless you believe that addition and subtraction is more difficult than differentiation.

\begin{align*}e^x &= 1 + x +\tfrac12x^2 + \mathcal O(x^3) \\ \implies e^{-x} &= 1 - x +\tfrac12x^2 + \mathcal O(x^3) \\[8pt] \frac{e^x + e^{-x} -2}{x^2} &= \frac{(1 + x +\tfrac12x^2) + (1 - x +\tfrac12x^2) -2}{x^2} \\ &= \frac{x^2 + \mathcal O(x^3) }{x^2} \\ &= 1 + \mathcal O(x) \\[12pt] \lim_{x \to 0} \frac{e^x + e^{-x} -2}{x^2} &= \lim_{x \to 0} \big( 1 + \mathcal O(x) \big) = 1 \end{align*}

Last edited by skipjack; March 11th, 2017 at 07:25 PM.

 March 11th, 2017, 07:26 PM #9 Global Moderator   Joined: Dec 2006 Posts: 17,919 Thanks: 1384 How do you know the power series for $e^x$?
 March 12th, 2017, 07:11 PM #10 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,574 Thanks: 931 Math Focus: Elementary mathematics and beyond $$\lim_{x\to0}\frac{e^x+e^{-x}-2}{x^2}=2\lim_{x\to0}\frac{\cosh x-1}{x^2}$$ Now use the Taylor series for $\cosh x$, $\sum_{n=0}^\infty\frac{x^{2n}}{(2n)!}$

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