I'm having some trouble solving this limit:

$$\lim_{x\to\infty} x\left[\left(\cosh x\right)^ \frac1x - \left(1+\frac1x\right)^x\right]$$

It's part of a set of limits wich should be solved using taylor. I tried this road:

$$\lim_{x\to\infty} x\left[e^{\frac1x\ln(\cosh x)}-e^{\ln\left(1+\frac1x\right)x}\right]$$

I then tried with algebraic manipulation using the definition of hyperbolic cosine $\frac12(e^x+e^{-x})$ and then I also played a bit with l'Hopital but it turns into something suspiciously compicated... I'm taking real analysis 1 and my toolset is:

-algebraic manipulation

-talyor series

-l'Hopital (if and only if all else fails)

I can't use more advanced techniques since they are not part of the course. I'm sure there's something obvious I'm missing. Any idea on how to proceed?

  • 1
    $\begingroup$ One thing to try is that your given limit equals $\lim_{y\rightarrow 0^+}\frac{\cosh(1/y)^y-(1+y)^{1/y}}{y}$. L'Hopital is appropriate for this fraction (by checking that both terms in the numerator have the same limit). $\endgroup$ – Michael Burr Jun 4 '16 at 13:21
  • $\begingroup$ @MichaelBurr thank you very much for your advice, however, being a beginner in analysis, I prefer to avoid l'Hopital whenever I can, since while it may give you results really fast at the same time it keeps you from noticing where your weaknesses are. $\endgroup$ – edo Jun 4 '16 at 18:51

I have tried several approaches, and the fastest seems the following [it does not require L'Hopital]:

$$ \lim_{x\to\infty} x[e^{\ln(\cosh x)\frac1x}-e^{\ln(1+\frac1x)x}]=\lim_{x\to\infty} e^{\ln x}[e^{\ln(\cosh x)\frac1x}-e^{\ln(1+\frac1x)x}] $$ $$ \lim_{x\to\infty} e^{\phi_1(x)}-e^{\phi_2(x)}\ . $$ Let us analyze the series expansions of the two functions in the exponent up to second next-to-leading order $$ \phi_1(x)=\ln x+\frac{1}{x}\ln(\cosh(x))=\ln x+\frac{1}{x}\ln\left(\frac{1}{2}(e^x+e^{-x})\right)=\ln x-\frac{\ln 2}{x}+1+\frac{1}{x}\ln(1+e^{-2x}) $$ $$ \sim \ln x+1-\frac{\ln 2}{x} $$ $$ \phi_2(x)=\ln x+x \ln(1+1/x)\sim\ln x +1-\frac{1}{2x} $$ Therefore, Taylor-expanding the exponentials [of the type $e^{-c/x}$] around $x=\infty$ $$ e^{\phi_1(x)}\sim e x e^{-\ln 2/x}\sim e x -e \ln (2)+\frac{e \ln ^2(2)}{2 x}-\frac{e \ln ^3(2)}{6 x^2}+\ldots $$ $$ e^{\phi_2(x)}\sim e x-\frac{e}{2}+\frac{e}{8 x}-\frac{e}{48 x^2}+\ldots $$ we get the final result by subtracting one from the other $$ \boxed{\lim_{x\to\infty} x[e^{\ln(\cosh x)\frac1x}-e^{\ln(1+\frac1x)x}]=\frac{e}{2}-e\ln 2\approx -0.525028...} $$

  • $\begingroup$ thanx a lot, it took me a minute to understand why you stated this: $$\ln x-\frac{\ln 2}{x}+1+\frac{1}{x}\ln(1+e^{-2x})\sim \ln x+1-\frac{\ln 2}{x}$$ and that was the root of my problem: $\\$we can ignore $~~$ $\frac{1}{x}\ln\left(1+e^{-2x}\right)$ $~~$ because it's approaching zero "way faster" than$~~$ $-\frac{\ln 2}{x}$. $\\$ I guess I'm still a bit shaky on when to use asymptotic equalities. Thanx for the editing as well, I'm new to MathJax. All the best! $\endgroup$ – edo Jun 4 '16 at 19:54
  • $\begingroup$ You're welcome. Very challenging limit indeed...and remember that limits and asymptotic expansions can only be learnt by the combination of three tools: i) PRACTICE ii) PRACTICE, and I let you figure out the third... $\endgroup$ – Pierpaolo Vivo Jun 4 '16 at 20:01

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