I only managed to show that:


I don't know how to bound it better, or maybe there is so simpler way?

  • $\begingroup$ hy don't you use L'Hospital? $\endgroup$ – Dr. Sonnhard Graubner May 17 '17 at 15:13
  • $\begingroup$ the result should be $6\sqrt{6}$ $\endgroup$ – Dr. Sonnhard Graubner May 17 '17 at 15:13
  • $\begingroup$ @Dr.SonnhardGraubner That is exactly what i found. $\endgroup$ – hamam_Abdallah May 17 '17 at 19:08

You can apply L'Hospital's rule to limits of this kind; you just have to be a bit clever and introduce a logarithm:

$$\begin{align} \lim_{x\to\infty}\left(\frac{2^{\frac{1}{x}}+3^{\frac{1}{x}}}{2}\right)^{3x} &= \lim_{x\to\infty}\exp\ln\left(\frac{2^{\frac{1}{x}}+3^{\frac{1}{x}}}{2}\right)^{3x}\\ &= \exp\lim_{x\to\infty}\ln\left(\frac{2^{\frac{1}{x}}+3^{\frac{1}{x}}}{2}\right)^{3x}\\ &= \exp\lim_{x\to\infty}\left(3x\cdot\ln\left(\frac{2^{\frac{1}{x}}+3^{\frac{1}{x}}}{2}\right)\right)\\ &= \exp\lim_{x\to\infty}\frac{3\ln\left(\frac{2^{\frac{1}{x}}+3^{\frac{1}{x}}}{2}\right)}{x^{-1}} \end{align}$$

That last limit is of the form $\frac00$, so it's ready for an application of L'Hospital's rule. Can you take it from there?

  • $\begingroup$ Yup, I think I did this. $\endgroup$ – UfmdFkiF May 17 '17 at 15:33
  • $\begingroup$ @TheMeff With first order Taylor expansion is more elegant. $\endgroup$ – hamam_Abdallah May 17 '17 at 19:10


$$2^{\frac {1}{x}}=e^{\frac {1}{x}\ln 2}$$

$$=1+\frac {\ln 2}{x}(1+\epsilon_2 (x)) $$

by the same for $3$, the function becomes

$$e^{3x\ln (1+\frac {\ln 6}{2x}(1+\epsilon (x))}$$

using fact that $$\ln (1+X)\sim X \;\;(X\to 0) ,$$

we get the limit $$ 6^{\frac {3}{2}} $$


As it is who the power infinity form we subtract the (1/×)=t

=e^(lim t tends zero [(2^t+3^t-2)/2])×(3/t)

Lim x tends zero( e^x -1)/x =ln(e ) .so (e) replace in above question with 2 and 3 . so you get= e^[ 3/2 (ln2 +ln3 )] = e^[ln (6)^3/2]


  • 1
    $\begingroup$ Welcome to StackExchange! Please read this tutorial and use it to typeset your Maths nicely so people can follow what you are saying easily $\endgroup$ – lioness99a May 18 '17 at 13:10

Compute first the limit of the (natural) logarithm and do the substitution $x=1/t$: $$ \lim_{t\to0^+}\log\left(\left(\frac{2^t+3^t}{2}\right)^{1/t}\right)= \lim_{t\to0^+}\frac{\log(2^t+3^t)-\log2}{t} $$ This is the derivative at $0$ of $f(t)=\log(2^t+3^t)$; since $$ f'(t)=\frac{2^t\log 2+3^t\log 3}{2^t+3^t} $$ the limit is $\frac{\log 2+\log3}{2}=\log\sqrt{6}$.

Hence the original limit is $\exp(\log\sqrt{6})=\sqrt{6}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.