Showing that $\lim_{x\to 0}\frac{\ln(\sin x+1)}{\sin x} = 1$ How can I show that $\lim_{x\to 0}\frac{\ln(\sin x+1)}{\sin x} = 1$?
I think that this function is continuous (maybe even uniform continuous) in all $R$ but $0$.
But because it is undefined at $x=0$ I'm not sure what I can do here.
Is there a known way for finding limits in situations like this? I thought of using the squeezing theorem somehow but couldn't find a way.
 A: Since 
$\lim_{x\rightarrow 0} \ln (\sin x +1)=0$ and $\lim_{x \rightarrow 0} \sin(x) = 0$,
By L'Hôpital's rule,
$$\lim_{x\rightarrow 0}\frac{\ln (\sin x + 1)}{\sin x}=\lim_{x\rightarrow 0}\frac{\cos x}{(\sin x +1)\cos x}=1$$
A: Call $\sin x=u$ and then rewrite the limit as $\ln\left ( \lim_{u\to0}(1+u)^{1/u}\right )$ and that inside limit is just $e$ so $\ln e=1$
A: Since plugging $0$ into the limit would result in an expressions of the form $\frac{0}{0}$, we can apply L'Hôpital's rule to the expression.
\begin{align}
\lim\limits_{x\rightarrow0}\frac{\ln(\sin x + 1)}{\sin(x)} &= \lim\limits_{x\rightarrow0}\frac{1}{\sin x + 1}\cdot\frac{\cos x}{\cos x}\\
&=\lim\limits_{x\rightarrow0}\frac{1}{\sin x + 1}\\
&= 1
\end{align}
A: Substituting $\sin x=t$,
$$\lim_{x\to 0}\frac{\ln(\sin x+1)}{\sin x} = \lim_{t\to 0}\frac{\ln(t+1)}t $$ which is known to be $1$.
A: Taylor expand twice, first order actually suffices.
For $x\sim 0$ we have $$\sin x\sim 0$$ and for $x\sim 0$ we have
$$
\ln(1+x)\sim x
$$
Making your limit pretty easy
$$
\lim_{x\to 0}\frac{\ln(\sin x+1)}{\sin x}=\lim_{x\to 0}\frac{x}{ x}=1  
$$
A: Although somewhat cumbersome, consider the following argument:
First note that for $x \to 0$ we have $\sin(x) \to 0$. The series expansion of the natural logarithm around $1$ is given by $\log(1 + x) = x + \mathcal{O}(x^2)$. For sin we have $\sin(x) = x + \mathcal{O}(x^3)$ Thus
\begin{align}
  \lim_{x \to 0} \frac{\log(\sin(x) + 1)}{\sin(x)} &= \lim_{x \to 0} \frac{x + \mathcal{O}(x^2)}{x + \mathcal{O}(x^3)} \\
  &= \lim_{x \to 0} \frac{1 + \mathcal{O}(x)}{1 + \mathcal{O}(x^2)} \\
  &= 1
\end{align}
