$\lim \limits_{x\to 0} \frac {4^{\sin x}-2^{\tan 2x}}{3^{\sin 2x}-9^{\tan x}}=?$

Question

find the limit :

$$\lim_{x\to 0} \frac {4^{\sin x}-2^{\tan 2x}}{3^{\sin 2x}-9^{\tan x}}=?$$

my try :

$$\lim_{x\to 0} \frac {2^{2\sin x}-2^{\tan 2x}}{3^{\sin 2x}-3^{2\tan x}}=$$

$$\lim_{x\to 0} \frac {2^{2\sin x}-2^{\frac{\sin 2x}{\cos2x}}}{3^{\sin 2x}-3^{\frac{2\sin x}{cosx}}}=$$

$$\lim_{x\to 0} \frac {2^{2\sin x}-2^{\sin 2x}+2^{\cos 2x}}{3^{\sin 2x}-3^{2\sin x}+3^{\cos x}}=?$$

now ?

Answer 1

Hint. One may write, by using a Taylor series expansion, as $x \to 0$, $$ \begin{align} \frac {4^{\sin x}-2^{\tan 2x}}{3^{\sin 2x}-9^{\tan x}}&= \frac {2^{\tan 2x}\cdot \left(2^{2\sin x-\tan 2x}-1\right)}{3^{2\tan x}\cdot \left(3^{\sin 2x-2\tan x}-1\right)}=\frac {2^{\tan 2x}\cdot \left(-x^3 \ln 8+o(x^3)\right)}{3^{2\tan x}\cdot \left(-x^3 \ln 9+o(x^3)\right)} \to \frac{\ln 8}{\ln 9}. \end{align} $$

Remark. Observe that, in general, $$ \frac{a^m}{a^n}=a^{m-n}\neq a^{\frac{m}{n}}. $$

Answer 2

$$\lim_{x\to0}\frac {2^{\tan 2x}\left(2^{(2\sin x-\tan 2x)}-1\right)}{3^{(2\tan x} \left(3^{(\sin 2x-2\tan x)}-1\right)}$$ $$=\lim_{x\to0}\frac {2^{\tan 2x}}{3^{2\tan x}}\cdot\underbrace{\lim_{x\to0}\dfrac{2^{(2\sin x-\tan 2x)}-1}{2\sin x-\tan 2x}}_{(1)}\cdot\lim_{x\to0}\dfrac1{\underbrace{\dfrac{3^{(\sin 2x-2\tan x)}-1}{\sin 2x-2\tan x}}_{(2)}}\cdot\lim_{x\to0}\dfrac{2\sin x-\tan 2x}{\sin 2x-2\tan x}$$

For $(1),(2)$ use $\lim_{h\to0}\dfrac{a^h-1}h=\ln a$

Now $\lim_{x\to0}\dfrac{2\sin x-\tan 2x}{\sin 2x-2\tan x}$

$=\lim_{x\to0}\dfrac{\cos x}{\cos2x}\cdot\lim_{x\to0}\dfrac{2\sin x\cos2x-\sin2x}{\sin2x\cos x-2\sin x}$

$=\lim_{x\to0}\dfrac{2\sin x(\cos2x-\cos x)}{2\sin x(\cos^2x-1)}$

$=2\cdot\dfrac32\cdot\dfrac12\lim_{x\to0}\dfrac{\sin\dfrac{3x}2}{\dfrac{3x}2}\cdot\lim_{x\to0}\dfrac{\sin\dfrac x2}{\dfrac x2}\cdot\dfrac1{\left(\lim_{x\to0}\dfrac{\sin x}x\right)^2}=?$