<s>[INST] How to solve this limit involving sine and log? I've tried L'Hopital's Rule but the differentiated numerator involves cos(1/x) which does not exist when x approaches 0.
$$ \lim_{x\to 0^+} \frac{x^2sin\frac{1}{x}}{\ln(1+2x)}$$
 [/INST] We know
$$\lim_{x\rightarrow 0} \ x\sin\left(\frac{1}{x}\right) = 0$$
$$\ln(1+2x) = 2x - \dfrac{4x^2}{2} + \dfrac{8x^3}{3} ... $$
So the limit reduces to :
$$\lim_{x\rightarrow 0} \ \frac{x\sin\left(\frac{1}{x}\right)x}{2x - \dfrac{4x^2}{2} + \dfrac{8x^3}{3} ..}$$
$$ = \lim_{x\rightarrow 0} \ \frac{x\sin\left(\frac{1}{x}\right)}{2 - \dfrac{4x}{2} + \dfrac{8x^2}{3} ..} = 0$$
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