I found the following problem in my calculus book:

$$\lim_{x\to \infty } \left(\frac{\ln (2 x)}{\ln (x)}\right)^{\ln (x)} $$

I tried to solve it using log rules and l'Hôpital's rule with no success, can someone give me any hints on how to go about this?


HINT: Using Sum of Logarithms,

$$\dfrac{\ln2x}{\ln x}=1+\dfrac{\ln2}{\ln x}$$

and set $\dfrac{\ln2}{\ln x}=n$ in $$\lim_{n\to\infty}\left(1+\dfrac1n\right)^n=e$$

Finally if $a^x=M$ from the definition, $x=\log_aM$

$$\displaystyle\implies a^{\log_aM}=M$$ (See also)

  • $\begingroup$ @BrianJ, Do you understand the difference between your edition & my original answer? $\endgroup$ – lab bhattacharjee Jul 31 '15 at 4:49
  • $\begingroup$ I based it around what looked like substituting the exponent for n. So I updated the others to match. If you did more than a simple substitution, then I clearly missed it. $\endgroup$ – Brian J Jul 31 '15 at 4:55

$$\lim_{x\to \infty }\left(\frac{\ln(2x)}{\ln x}\right)^{\ln x}=\lim_{x\to\infty}\left(\frac{\ln2+\ln x}{\ln x}\right)^{\ln x}= \lim_{x\to\infty}\left(1+\frac{1}{\frac{\ln x}{\ln 2}}\right)^{\frac{\ln x}{\ln 2}\cdot\ln 2}=e^{\ln 2}=2$$


$$\lim_{x\to \infty } \Big(\frac{\ln(2x)}{\ln(x)}\Big)^{\ln(x)} $$ $$=\lim_{x\to \infty } \Big(\frac{\ln(x)+\ln 2}{\ln(x)}\Big)^{\ln(x)} $$ $$=\lim_{x\to \infty } \Big(1+\frac{\ln(2)}{\ln(x)}\Big)^{\ln(x)} $$ $$=\lim_{x\to \infty }exp\Big(\ln(x)\Big(1+\frac{\ln(2)}{\ln(x)}-1\Big)\Big) $$ $$=\lim_{x\to \infty }exp\Big(\ln (x)\frac{\ln(2)}{\ln(x)}\Big)\Big) $$ $$=\lim_{x\to \infty }exp\Big(\ln 2\Big)=e^{\ln 2}=2 $$

  • $\begingroup$ How do you reach from third expression to fourth? I think fourth expression should be $$\lim_{x \to \infty}\exp\left\{\ln x\ln\left(1 + \frac{\ln 2}{\ln x}\right)\right\}$$ $\endgroup$ – Paramanand Singh Aug 11 '15 at 9:14
  • $\begingroup$ Kindly, notice $$\lim_{x\to \infty}\left(1+\frac{1}{x}\right)^{x}=\lim_{x\to \infty}e^{x\left(1+\frac{1}{x}-1\right)}=e$$ $\endgroup$ – Harish Chandra Rajpoot Aug 11 '15 at 9:43
  • $\begingroup$ No it happens like $$\lim_{x\to \infty}\left(1+\frac{1}{x}\right)^{x}=\lim_{x\to \infty}e^{x\log\left(1+\frac{1}{x}\right)}=\exp\left\{\lim_{x \to \infty}x\log\left(1+\frac{1}{x}\right)\right\} = \exp(1) = e$$ $\endgroup$ – Paramanand Singh Aug 11 '15 at 9:50
  • $\begingroup$ Yes, it's correct method. But I showed you a short cut method. $\endgroup$ – Harish Chandra Rajpoot Aug 11 '15 at 9:59
  • $\begingroup$ It appears that you are using the following crafty rule: if $f(x) \to 1, g(x) \to \infty$ then $\{f(x)\}^{g(x)} \to \exp\{\lim g(x)(f(x) - 1)\}$ $\endgroup$ – Paramanand Singh Aug 11 '15 at 10:29

$$\lim\limits_{x\to \infty} \left(\frac{\ln (2x)}{\ln x}\right)^{\ln x} $$ $$=\lim\limits_{x\to \infty} \left(\frac{\ln 2+\ln x}{\ln x}\right)^{\ln x} $$ $$=\lim\limits_{x\to \infty} \left(\frac{\ln 2}{\ln x}+1\right)^{\ln x} $$ $$=\lim\limits_{x\to \infty} \left(\frac{1}{\frac{\ln x}{\ln 2}}+1\right)^{\ln x} $$ Let $n=\frac{\ln x}{\ln 2}$, then $$\lim\limits_{n\to \infty} \left(1+\frac{1}{n}\right)^{n\ln 2} $$ $$=\left(\lim\limits_{n\to \infty} \left(1+\frac{1}{n}\right)^n\right)^{\ln 2} $$ $$=e^{\ln 2}=2$$


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