# Can't solve a difficult limit

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?

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

Hint:

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 $$\frac{\cos x\ln x}{\ln(1+\cosh x)}\to 0?$$

• I understood! Thank you very much! – 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$$\lim_{x\to\infty}\frac{\left(\sqrt{x^2+x}-x\right)+\frac{\cos(x)\log(x)}x}{\frac{\log(\cosh x)}x}.$$

• Thanks for the hint! – 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

• $$\ln(1+\cosh(x))\sim x$$

• $$\sqrt{x^2 + x} - x\sim \frac12$$

therefore

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

• 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. – Paramanand Singh Nov 12 '19 at 4:58
• @ParamanandSingh Thanks I fix that! – user Nov 12 '19 at 8:17
• And if I am not wrong you were using the name gimusi earlier. It took me a while to figure this out :) – Paramanand Singh Nov 12 '19 at 9:58
• @ParamanandSingh Yes your guess is right! – user Nov 12 '19 at 10:23