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I'm trying to solve this limit:

$$\lim\limits_{x \to 0}{\frac{\cosh x - 1}{x^2}}$$ where $\cosh x = \frac{e^x + e^{-x}}{2}$

$\lim\limits_{x \to 0}{\frac{\cosh x - 1}{x^2}} = $ $\lim\limits_{x \to 0}{\frac{\frac{e^x + e^{-x}}{2} - 1}{x^2}} = $ $\lim\limits_{x \to 0}{\frac{\frac{e^x + e^{-x} - 2}{2}}{x^2}} = $ $\frac{1}{2}\lim\limits_{x \to 0}{\frac{e^x + e^{-x} - 2}{x^2}} = $ $\frac{1}{2}\lim\limits_{x \to 0}{\frac{(e^x - 1) + (e^{-x} - 1)}{x^2}} = $ $\frac{1}{2}\lim\limits_{x \to 0}{\frac{1}{x}\frac{(e^x - 1) + (e^{-x} - 1)}{x}} = $ $\frac{1}{2}\lim\limits_{x \to 0}{\frac{1}{x}(\frac{e^x - 1}{x} + \frac{e^{-x} - 1}{x})} = $ $\frac{1}{2}\lim\limits_{x \to 0}{\frac{1}{x}(\frac{e^x - 1}{x} - \frac{e^{-x} - 1}{-x})} = ???$

I'm pretty sure I'm close to the answer, because I think I can put $\lim\limits_{x \to 0}{\frac{e^x - 1}{x}} = 1$ to use somehow to finish it off. That $\frac{1}{x}$ is quite annoying. :)

Any hints?

5 Answers 5

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$$\dfrac{\cosh x-1}{x^2}=\dfrac{2\cosh x-2}{2x^2}=\dfrac{e^x+e^{-x}-2}{2x^2}=\dfrac{(e^x-1)^2}{2x^2e^x}=\dfrac1{2e^x}\left(\dfrac{e^x-1}x\right)^2$$

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    This reminds me of $\cosh x - 1=2\sinh^{2}(x/2)$ +1 – 2017-02-23
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You may wish to consider the series expansion for $\cosh x$ is $$\frac{e^x+e^{-x}}{2}=1+\frac{x^2}{2!}+\frac{x^4}{4!}+\frac{x^6}{6!}+\dots$$ So the limit is $\frac{1}{2}$.

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    Too advanced for my level. (Keep it for others' sakes, though.) – 2017-02-23
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Use L'Hopital's rule, $\cosh'=\sinh$, $\sinh'=\cosh$, $\cosh(0)=1$ and $\sinh(0)=0$. This gives $$\lim_{x\rightarrow0}\frac{\cosh(x)-1}{x^2}=\lim_{x\rightarrow0}\frac{\sinh(x)}{2x}=\lim_{x\rightarrow0}\frac{\cosh}{2}=\frac{1}{2}$$

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    See the tag please. – 2017-02-23
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    @Rohan I know, OP has added it after I typed my answer (see his comment on Maczinga's answer). I know this is another question, but what is the point of calculating limits without L'Hopital's. Pedagogical? And if so, why not to come up with a problem that does not admit use of L'Hopital's? – 2017-02-23
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Ok, if you want not to use L'Hospital rule then consider the developpement for $e^x=\sum_{n=0}^\infty\frac{x^n}{n!}$. Stopping at the term in $x^2$ we have: $1+x+1/2x^2$ and for $e^{-x}$ we have $1-x+1/2x^2$. Hence

$\frac1{2}\cdot\lim\limits_{x \to 0}{\frac{e^x + e^{-x} - 2}{x^2}} = $ $\frac1{2}\cdot\lim\limits_{x \to 0}{\frac{1+x+1/2x^2+1-x+1/2x^2 - 2}{x^2}} = $ $\frac1{2}\cdot\lim\limits_{x \to 0}{\frac{x^2}{x^2}}$

Now you can keep from here.

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    Sorry, forgot to add `limits-without-lhopital`. – 2017-02-23
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    ah that'changes everything – 2017-02-23
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    My apologies for not including it originally. :) – 2017-02-23
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$(e^x -1)(e^{-x}-1) = 2 - e^x - e^{-x}$ and hence $$\lim_{x \rightarrow 0} \frac{e^x + e^{-x}-2}{2x^2} = \lim_{x \rightarrow 0} \frac{-(e^x -1)(e^{-x}-1)}{2x^2} = \frac{1}{2}\lim_{x \rightarrow 0} \frac{e^{x}-1}{x} \lim_{x \rightarrow 0} \frac{e^{-x}-1}{-x} = \frac{1}{2}$$