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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 ?

Almot1960
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  • Put $x = 0$ to the last expression that you got and you'll get it to be $2/3$. – rookie Feb 18 '17 at 11:37
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    But that can't be true since the OP hasn't simplified the expression correctly. – Nilabro Saha Feb 18 '17 at 11:39
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    Check your expansion of $2^ {\frac{\sin(2x)}{\cos(2x)}} $ in the numerator and for $3^{\frac{2\sin(x)}{\cos(x)}}$ in the denominator. –  Feb 18 '17 at 12:16

2 Answers2

2

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, $$ \frac{a^m}{a^n}=a^{m-n}\neq a^{\frac{m}{n}}. $$

Olivier Oloa
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    why ?$\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} $?? – Almot1960 Feb 18 '17 at 12:01
  • @Almot1960 What happens to $\tan(x)$ ( and $\tan(2x)$ ) when $x \to 0$ ??? –  Feb 18 '17 at 12:20
  • @Almot1960 Because, as $x \to 0$, one has $\frac {2^{\tan 2x}\cdot \left(-x^3 \ln 8+o(x^3)\right)}{3^{2\tan x}\cdot \left(-x^3 \ln 9+o(x^3)\right)} \sim \frac {2^{0}\cdot \left(-x^3 \ln 8\right)}{3^{0}\cdot \left(-x^3 \ln 9\right)}=\frac{\ln 8}{\ln 9}.$ – Olivier Oloa Feb 18 '17 at 12:21
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 $\lim_{x\to0}\dfrac{2\sin x-\tan 2x}{\sin 2x-2\tan x}$

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

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

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

lab bhattacharjee
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