I need to solve this limit ${\lim_ {x\to {+∞}}}{\frac{{x}(\sqrt{x^2 + x} - x) +\cos(x)\ln(x)}{\ln(1+\cosh(x))}}$

I've tried to use Taylor's Theorem with Peano's Form of Remainder, but first time I forgot that ${x\to{+∞}}$, so I made a substitution ${t=\frac{1}{x}}$, then I just didn't get anything (I've got ${o({\frac{1}{t}})}$ (or ${o((t-1)^3)}$ and too complicated expression) which doesn't disappear). I've thought to use L'Hospital's rule, but there's a problem with defining indeterminate form. Here we have ${\cos(x)\ln(x)}$ that sometimes becomes ${0\cdot∞}$. Then I thought about the existence of this limit and... WolframAlpha says it doesn't exist. But the answer in my book is 1/2.

So now Ii don't know how to solve it or does it even exist or not. Can anyone give me at least a hint of how to solve this problem?

  • $\begingroup$ Hint: $f(x)/g(x)$ can sometimes be usefully written as $(f(x)/x)\times(x/g(x))$. $\endgroup$ – Barry Cipra Nov 11 '19 at 22:34


First the expression in two, and use equivalents:

We have $\cosh x\sim_{+\infty}\frac12\mathrm e^x$, so $1+\cosh x\sim \frac12\mathrm e^x$, and finally $$\ln(1+\cosh x)\sim_{+\infty}x-\ln 2\sim_{+\infty} x$$

. On the other hand, $$x(\sqrt{x^2 + x} - x)=\frac{x(\not x^2 + x - \not x^2)}{\sqrt{x^2 + x} + x}\sim_{+\infty}\frac{x^2}{2x}=\frac x2,$$ so that $$\frac{x(\sqrt{x^2 + x} - x)}{\ln(1+\cosh x)}\sim_{+\infty}\frac{\frac12 x}{x}=\frac 12.$$ Can you show that $$\frac{\cos x\ln x}{\ln(1+\cosh x)}\to 0?$$

  • $\begingroup$ I understood! Thank you very much! $\endgroup$ – IPPK Nov 11 '19 at 23:13

Hint: Note that, for large $x$, $\cosh x$ behaves as $\frac12e^x$. So, $\lim_{x\to\infty}\frac{\log(\cosh x)}x=1$. And your limit is equal to$$\lim_{x\to\infty}\frac{\left(\sqrt{x^2+x}-x\right)+\frac{\cos(x)\log(x)}x}{\frac{\log(\cosh x)}x}.$$

  • $\begingroup$ Thanks for the hint! $\endgroup$ – IPPK Nov 11 '19 at 23:14

You can start with the definition of $\cosh x$ and note that $$\log(1+\cosh x) =\log(e^x+e^{-x} +2)-\log 2=x+\log(1+e^{-2x}+2e^{-x})-\log 2$$ and hence $\dfrac{\log(1+\cosh x)} {x} \to 1$ as $x\to\infty $. Therefore we can replace $\log(1+\cosh x) $ by $x$ in the denominator (this is basically multiplying the given expression by the ratio $(\log(1+\cosh x)) /x$).

Based on numerator we can split the fraction in two parts and since $(\log x) /x$ tends to $0$ the second part has limit $0$ (remember $\cos x$ is bounded).

The desired limit is thus equal to the limit of $(\sqrt{x^2+x}-x)$ which is $1/2$ (that's the easiest part). Thus the problem entirely rests on the fundamental limit $$\lim_{x\to\infty} \frac{\log x} {x} =1$$ and the rest of it just plain algebraic manipulation.


We have that

  • $\ln(1+\cosh(x))\sim x$

  • $\sqrt{x^2 + x} - x\sim \frac12$


$${\frac{{x}(\sqrt{x^2 + x} - x) +\cos(x)\ln(x)}{\ln(1+\cosh(x))}}\sim \frac12 + \cos(x)\frac{\ln(x)}{x}\to \frac12$$

  • $\begingroup$ There is a typo at the end. You need to write $(\ln x) /x$ instead of $\ln(x) x$. Just to clarify the downvote is not mine. $\endgroup$ – Paramanand Singh Nov 12 '19 at 4:58
  • $\begingroup$ @ParamanandSingh Thanks I fix that! $\endgroup$ – user Nov 12 '19 at 8:17
  • $\begingroup$ And if I am not wrong you were using the name gimusi earlier. It took me a while to figure this out :) $\endgroup$ – Paramanand Singh Nov 12 '19 at 9:58
  • $\begingroup$ @ParamanandSingh Yes your guess is right! $\endgroup$ – user Nov 12 '19 at 10:23

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