I found the following problem in my calculus book:
Solve:
$$\lim_{x\to \infty } \left(\frac{\ln (2 x)}{\ln (x)}\right)^{\ln (x)} $$
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?
$\endgroup$4 Answers
$\begingroup$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)
$\endgroup$ 2 $\begingroup$$$\lim_{x\to \infty }\left(\frac{\ln(2x)}{\ln x}\right)^{\ln x}=\lim_{x\to\infty}\left(\frac{\ln2+\ln x}{\ln x}\right)^{\ln x}= \lim_{x\to\infty}\left(1+\frac{1}{\frac{\ln x}{\ln 2}}\right)^{\frac{\ln x}{\ln 2}\cdot\ln 2}=e^{\ln 2}=2$$
$\endgroup$ $\begingroup$$$\lim_{x\to \infty } \Big(\frac{\ln(2x)}{\ln(x)}\Big)^{\ln(x)} $$ $$=\lim_{x\to \infty } \Big(\frac{\ln(x)+\ln 2}{\ln(x)}\Big)^{\ln(x)} $$ $$=\lim_{x\to \infty } \Big(1+\frac{\ln(2)}{\ln(x)}\Big)^{\ln(x)} $$ $$=\lim_{x\to \infty }exp\Big(\ln(x)\Big(1+\frac{\ln(2)}{\ln(x)}-1\Big)\Big) $$ $$=\lim_{x\to \infty }exp\Big(\ln (x)\frac{\ln(2)}{\ln(x)}\Big)\Big) $$ $$=\lim_{x\to \infty }exp\Big(\ln 2\Big)=e^{\ln 2}=2 $$
$\endgroup$ 6 $\begingroup$$$\lim\limits_{x\to \infty} \left(\frac{\ln (2x)}{\ln x}\right)^{\ln x} $$ $$=\lim\limits_{x\to \infty} \left(\frac{\ln 2+\ln x}{\ln x}\right)^{\ln x} $$ $$=\lim\limits_{x\to \infty} \left(\frac{\ln 2}{\ln x}+1\right)^{\ln x} $$ $$=\lim\limits_{x\to \infty} \left(\frac{1}{\frac{\ln x}{\ln 2}}+1\right)^{\ln x} $$ 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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