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I am asked to calculate the following limit $$ \lim_{x\to0}\frac{\ln(1+\sin x)}{\sin(2x)} $$

First, I tried expressing $1+\sin x=t$, then express $x$ from that equation but my equation seemed to just get more complicated, then I tried expressing the whole $\ln(x+1)=t$, but didn't get anywhere, I'm not allowed to use L'Hospital's Rule nor Taylor series, the answer seems to be $\Large \frac{1}{2}$.

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Notice, $$\lim_{x\to 0}\frac{\ln(1+\sin x)}{\sin(2x)}$$ $$=\lim_{x\to 0}\frac{\ln(1+\sin x)}{2\sin x\cos x}$$ $$=\frac 12\lim_{x\to 0}\frac{\ln(1+\sin x)}{\sin x}\cdot \frac{1}{\cos x}$$ let $\sin x=t\implies t\to 0$ as $x\to 0$, $$=\frac 12\left(\lim_{t\to 0}\frac{\ln(1+t)}{t}\right)\cdot \left(\lim_{x\to 0}\frac{1}{\cos x}\right)$$ $$=\frac 12(1)\cdot (1)=\color{red}{\frac 12}$$

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    $\begingroup$ Thanks, didn't realize (ln(1+t))/t was actually 1 $\endgroup$ – Bardh Mehmeti Jan 22 '16 at 19:26
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Hint: $\sin(2x) = 2\sin x\cos x$ and $\dfrac{\log(1+\sin x)}{\sin x} \to 1$

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Observe that $$\dfrac{\ln(1+\sin(x))}{\sin(2x)} = \dfrac{2x}{\sin(2x)}\left[\dfrac{\ln(1+\sin(x))}{x} \right]\left(\dfrac{1}{2}\right) = \left[\dfrac{\ln(1+\sin(x))}{x} \right]/\left[\sin(2x)/2x\right]\cdot \dfrac{1}{2}\text{.}$$ Now $$\dfrac{\ln(1+\sin(x))}{x} = \dfrac{\ln(1+\sin(x))-0}{x-0} = \dfrac{\ln(1+\sin(x))-\ln(1+\sin(0))}{x-0}$$ and $$\lim\limits_{x \to 0}\dfrac{\ln(1+\sin(x))}{x} = \lim_{x \to 0}\dfrac{\ln(1+\sin(x))-\ln(1+\sin(0))}{x-0} = [\ln(1+\sin(x))]^{\prime}_{x=0}\text{.}$$ The derivative of $\ln(1+\sin(x))$ is $$\dfrac{\cos(x)}{1+\sin(x)}$$ and evaluated at $x = 0$, we have $1$. Furthermore, $$\lim_{x \to 0}\dfrac{\sin(2x)}{2x} = 1$$ since as $x \to 0$, $2x \to 0$ as well.

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$$\lim\limits_{x \to 0}\frac{\ln(1+\sin(x))}{\sin(2x)}=\lim\limits_{x \to 0}\frac{\ln(1+\frac{\sin(x)}{x}x)}{\frac{\sin(2x)}{2x}2x}=\lim\limits_{x \to 0}\frac{\ln(1+x)}{2x}=\frac{1}{2}\lim\limits_{x \to 0}\frac{\ln(1+x)}{x}=\frac{1}{2}$$

We are using in between $\lim\limits_{x \to 0}\frac{\sin(x)}{x}=1$

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