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March 11th, 2017, 01:39 PM   #1
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Unhappy 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
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March 11th, 2017, 02:19 PM   #2
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Use L'Hôpital's rule.

Last edited by skipjack; March 11th, 2017 at 04:39 PM.
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March 11th, 2017, 02:25 PM   #3
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You need to do it twice.
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March 11th, 2017, 04:13 PM   #4
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This problem is perfect for Maclaurin series. Far easier than l'Hôpital too.
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March 11th, 2017, 06:14 PM   #5
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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 08:28 PM.
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March 11th, 2017, 07:10 PM   #6
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The limit is one.
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March 11th, 2017, 07:17 PM   #7
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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:

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March 11th, 2017, 08:18 PM   #8
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Quote:
Originally Posted by johng40 View Post
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 08:25 PM.
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March 11th, 2017, 08:26 PM   #9
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How do you know the power series for $e^x$?
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March 12th, 2017, 07:11 PM   #10
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$$\lim_{x\to0}\frac{e^x+e^{-x}-2}{x^2}=2\lim_{x\to0}\frac{\cosh x-1}{x^2}$$

Now use the Taylor series for $\cosh x$, $\sum_{n=0}^\infty\frac{x^{2n}}{(2n)!}$
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