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Problem

Evaluate the following using series expansion.

$$ \lim_{x \rightarrow 0} \frac{\cosh{x}-\cosh{2x}}{x \cdot \sinh{x} }$$


I really don't know what I should do. If I expand $\cosh$ and $\sinh$ normally, (they'd look something like $\sum_{n = 0}^{\infty}\frac{x^{2n}}{(2n)!}$) I get a fraction with two infinite sums with the limit $0$ on both top and bottom and if I expand it using Taylor, I have to use $x \neq 0$ and I get a sum with infinite summands.

I don't have anymore ideas and any help would be appreciated. Thank you in advance.

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$$\cosh(x)=1+\frac12x^2+\mathcal O(x^4)$$

$$\sinh(x)=x+\mathcal O(x^3)$$

$$\frac{\cosh(x)-\cosh(2x)}{x\sinh(x)}=\frac{-\frac32x^2+\mathcal O(x^4)}{x^2+\mathcal O(x^3)}=\frac{-\frac32+\mathcal O(x^2)}{1+\mathcal O(x^2)}\stackrel{x\to0}\longrightarrow\boxed{-\frac32}$$

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  • $\begingroup$ Thank you for your answer too. I think I understood it now. $\endgroup$
    – user396246
    May 22 '17 at 22:45
  • $\begingroup$ No problem! :-) $\endgroup$ May 22 '17 at 22:48
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You're correct to expand $\cosh$ and $\sinh$ normally, but since we're taking a limit near $0$, we can neglect all but the first few terms - I'd guess only the first two terms would matter here. Informally you could write $\cosh x = 1 + \frac{x^2}{2} + \dots$, or formally you could say $\cosh x = 1 + \frac{x^2}{2} + \mathcal{O}(x^4)$.

In this way, $$\begin{align} \lim_{x \to 0} \frac{\cosh x - \cosh 2x}{x \cdot \sinh x} &= \lim_{x \to 0} \frac{(1 + \frac{x^2}{2} + \mathcal{O}(x^4)) - (1 + 2x^2 + \mathcal{O}(x^4))}{x \cdot (x + \frac{x^3}{3!} + \mathcal{O}(x^5))} \\ &= \lim_{x \to 0} \frac{-\frac3 2 x^2 + \mathcal{O}(x^4)}{x^2 + \mathcal{O}(x^4)} \\ &= \lim_{x \to 0} \frac{-\frac3 2 + \mathcal{O}(x^2)}{1 + \mathcal{O}(x^2)} \\ &= -\frac{3}{2}\end{align}$$

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  • $\begingroup$ Ah, I get it now! Thank you very much! $\endgroup$
    – user396246
    May 22 '17 at 22:45
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As other answers said, you are correct and the limit is easy to find using this way.

However, the same procedure can be used to see how is approached the limit. Just use one extra term in the expansions $$\cosh(x)=1+\frac{x^2}{2}+\frac{x^4}{24}+O\left(x^6\right)$$ $$\cosh(2x)=1+2 x^2+\frac{2 x^4}{3}+O\left(x^6\right)$$ $$\sinh(x)=x+\frac{x^3}{6}+\frac{x^5}{120}+O\left(x^7\right)$$ $$\frac{\cosh(x)-\cosh(2x)}{x\sinh(x)}=\frac{\left(1+\frac{x^2}{2}+\frac{x^4}{24}+O\left(x^6\right) \right)-\left(1+2 x^2+\frac{2 x^4}{3}+O\left(x^6\right) \right) } { x^2+\frac{x^4}{6}+\frac{x^6}{120}+O\left(x^7\right)}$$ $$\frac{\cosh(x)-\cosh(2x)}{x\sinh(x)}=\frac{-\frac{3 x^2}{2}-\frac{5 x^4}{8}+O\left(x^6\right) }{ x^2+\frac{x^4}{6}+\frac{x^6}{120}+O\left(x^7\right) }$$ Simplify by $x^2$ and use long division to finally get $$\frac{\cosh(x)-\cosh(2x)}{x\sinh(x)}=-\frac{3}{2}-\frac{3 x^2}{8}+O\left(x^4\right)$$

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