Here it is:
$$\lim_{x\to 0} \frac{6^{2x}-7^{-2x}}{\sin 3x - 2x}$$
So, thats what I've done:
$$=\lim_{x\to0} \frac{(36^{x}-1)-(49^{-x}-1)}{\sin 3x - 2x}$$
$$=\frac {x(\ln 36+\ln 49)} {\lim_{x\to0}{\sin 3x} - \lim_{x\to0}{2x}}$$ $$=\frac {x(\ln 36+\ln 49)} {3x-0}$$
The answer I've got is $\frac{\ln(1764)}{3}$ but wolframalpha says the answer is just $\ln(1764)$ without 3 in denominator. What did I do wrong ?
$$\lim_{x\to0}\frac{6^{2x}-7^{-2x}}{\sin3x-2x}$$
$$=\lim_{x\to0}\frac{\dfrac{36^x-1}x-\dfrac{(7^{-2})^x-1}x}{3\cdot\dfrac{\sin3x}{3x}-2}$$
Now for $a=e^{\ln(a)}$, $\lim_{h\to0}\dfrac{a^h-1}h=\ln a\cdot\lim_{h\to0}\dfrac{e^{h\ln a}-1}{h\ln a}=\ln(a)$