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Trying to compute

$$\lim_{x \to \infty}\left(\;\cosh \sqrt{x+1}-\cosh{\sqrt{x}}\;\right)^{1/\sqrt{x}}=e$$

I arrived to the equivalent expression $$\lim_{x \to \infty}\frac{1}{\sqrt{x}}\log\left(\cosh \sqrt{x+1}-\cosh{\sqrt{x}} \right)$$but couldn't undo the indeterminate form $0 \cdot \infty$. L'Hôpital seems to become a loop.

Any suggestions?

Thanks in advance.

Blue
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Tavasanis
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    Can you please let me know about the source of this question by editing your post? Also, are you aware of the definition of the $\cosh$ function? This is for the purposes of adding context, see here for details. – Sarvesh Ravichandran Iyer Jul 12 '21 at 06:53
  • This question comes from a list of benchmarking tests of limits on different CAS systems, many of which do not show any answer. Second, $\cosh(x)=(e^x+e^{-x})/2$, and I used such a definition in my attempts. – Tavasanis Jul 12 '21 at 06:58
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    Thank you very much. You may add such content to your question post as well. Also remember this : if you want to mention your attempt, then you are better mentioning mathematical details rather than an explanation (e.g. this method has loopholes). If you feel that your attempt doesn't need feedback or could be irrelevant, then details on your source and background take precedence in the form of context, as they should do. Regularly mentioning your source (in as much detail as possible e.g. references and textbooks) and definitions on this site are the best forms of context. – Sarvesh Ravichandran Iyer Jul 12 '21 at 07:00

3 Answers3

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Let's take a closer look at \begin{align} 2(\cosh(\sqrt{x+1})-\cosh(\sqrt{x}))&= e^{\sqrt{x+1}}+e^{-\sqrt{x+1}} -e^{\sqrt{x}}-e^{-\sqrt{x}}\\ &\sim_{x \infty} e^{\sqrt{x+1}} - e^{\sqrt{x}}\\ &\sim_{x \infty} e^{\sqrt{x}}(-1+e^{\sqrt{x+1}-\sqrt{x}})\\ &\sim_{x \infty} e^{\sqrt{x}}(-1+e^{\sqrt{x}(\sqrt{1+1/x}-1)})\\ &\sim_{x \infty} e^{\sqrt{x}}(-1+e^{1/(2\sqrt{x})})\\ &\sim_{x \infty} \frac{e^{\sqrt{x}}}{2\sqrt{x}}.\\ \end{align} Since $\frac{-1}{\sqrt{x}}\log(4\sqrt{x}) \to 0$, we have $$ \frac{1}{\sqrt{x}}\log\left(\cosh \sqrt{x+1}-\cosh{\sqrt{x}} \right) \sim \frac{1}{\sqrt{x}}\sqrt{x}=1, $$ and by applying $e^\cdot$ $$ \lim_{x \to \infty}\left(\cosh \sqrt{x+1}-\cosh{\sqrt{x}} \right)^{\frac{1}{\sqrt{x}}}=e. $$

Michelle
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$$\lim_{x \to \infty}\frac{1}{\sqrt{x}}\log\left(\cosh \sqrt{x+1}-\cosh{\sqrt{x}} \right) = \lim_{x \to \infty} \frac{1}{\sqrt{x}}\log\left\{\frac{1}2\left(e^{\sqrt{x+1}}+e^{-\sqrt{x+1}}+e^{\sqrt{x}}+e^{-\sqrt{x+1}}\right)\right\}$$

It's easy to see that the contribution of the addition of the exponentials will be insignificant (for instance, divide by the limit without them and apply L'Hop to get $1$, or simply bound them by $1$ and use inequalities to see the growth is neglible when divided by $\sqrt{x}$. Hence we write this

$$\lim_{x \to \infty} \frac{1}{\sqrt{x}}\log\left\{\frac{1}2\left(e^{\sqrt{x+1}}+e^{\sqrt{x}}\right)\right\}$$

Again, comparing this to the same expression with the ${x+1}$ replaced with an $x$, it's easy to bound the result as say a constant time the exact argument of the logarithm, which will be negligible when divided out by the $\sqrt{x}$. So this limit is just

$$ \lim_{x \to \infty} \frac{1}{\sqrt{x}}\log\left\{\frac{1}2\left(e^{\sqrt{x}}+e^{\sqrt{x}}\right)\right\} = \lim_{x\to\infty} \frac{1}{\sqrt{x}}\log\left\{e^{\sqrt{x}}\right\} = 1. $$

Which readily yields the desired result

$$ \lim_{x \to \infty}\left(\cosh \sqrt{x+1}-\cosh{\sqrt{x}} \right)^{\frac{1}{\sqrt{x}}}=e$$

Cade Reinberger
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  • Excellent answer, thank you very much. I can't help wondering what theorem or result supports such bounding. – Tavasanis Jul 12 '21 at 07:05
  • They can all be done here with just calculus if one takes the effort to work out every step. For instance, for the second step, start with $\sqrt{x+1} \leq \sqrt{x}+1$, so $e^{\sqrt{x+1}} \leq e \cdot e^{\sqrt{x}}$, so the whole argument of the log is changed at most say a factor of $e$, which has the same limit since the added factor of $e$ goes to $0$, meaning the limits are equal by the squeeze theorem. More generally though, asymptotic arithmetic, as in @Michelle's answer, can simplify these kinds of calculations. – Cade Reinberger Jul 12 '21 at 07:12
  • That's OK. I see. Thank you again. – Tavasanis Jul 12 '21 at 07:14
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$$A=\cosh \left(\sqrt{x+1}\right)-\cosh \left(\sqrt{x}\right)=2 \sinh \left(\frac{\sqrt{x+1}-\sqrt{x}}{2} \right) \sinh \left(\frac{\sqrt{x}+\sqrt{x+1}}{2} \right)$$ When $x$ is large, using equivalents, $$\sinh \left(\frac{\sqrt{x+1}-\sqrt{x}}{2} \right)\sim\frac 1{4 \sqrt x}$$ $$\sinh\left(\frac{\sqrt{x}+\sqrt{x+1}}{2} \right)\sim \sinh(\sqrt{x})\sim \frac 12e^{\sqrt x}$$

$$A\sim 2 \times \frac 1{4 \sqrt x}\times \frac 12e^{\sqrt x}=\frac{e^{\sqrt{x}}}{4 \sqrt{x}}$$ $$B=A^{\frac 1{\sqrt x}}\sim \Bigg[\frac{e^{\sqrt{x}}}{4 \sqrt{x}} \Bigg]^{\frac 1{\sqrt x}}$$ Take logarithms $$\log(B)=\frac 1{\sqrt x}\log\Bigg[\frac{e^{\sqrt{x}}}{4 \sqrt{x}} \Bigg]=1-\frac 1{\sqrt x}\log(4\sqrt x)$$ and $B=e^{\log(B)}$

Claude Leibovici
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