Question: I am trying to compute the following limit without L'Hôpital's rule : $$L=\lim\_{x\to 0}\frac{\ln(\tan(x)+1)-\sin(x)}{x\sin(x)}$$

I evaluated ths limit using L'Hôpital's and found $-\frac12$ as the answer. 

I eventually ended up to : 
$$L=\frac12\lim\_{x\to 0} 3\cos^2(x)\sin(x)-\frac12$$
I find L'hopital's to be very lengthy in this case.

Is there another way to do it ? I'm stuck 

Thanks for the help, 
T.D.

Answer: 
Why not to compose Taylor series from the beginning
$$\frac{\log(\tan(x)+1)-\sin(x)}{x\sin(x)}$$
knowing that is must be at least to $O(x^2)$ because of the denominator
$$\tan(x)+1=1+x+\frac{x^3}{3}+O\left(x^5\right)$$
$$\log(\tan(x)+1)=x-\frac{x^2}{2}+\frac{2 x^3}{3}-\frac{7 x^4}{12}+O\left(x^5\right)$$
$$\log(\tan(x)+1)-\sin(x)=-\frac{x^2}{2}+\frac{5 x^3}{6}-\frac{7 x^4}{12}+O\left(x^5\right)$$
$$\frac{\log(\tan(x)+1)-\sin(x)}{x\sin(x)}=\frac{-\frac{x^2}{2}+\frac{5 x^3}{6}-\frac{7 x^4}{12}+O\left(x^5\right) } {x^2-\frac{x^4}{6}+O\left(x^5\right) }$$
Now, long division
$$\frac{\log(\tan(x)+1)-\sin(x)}{x\sin(x)}=-\frac{1}{2}+\frac{5 x}{6}-\frac{2 x^2}{3}+O\left(x^3\right)$$ which shows the limit and how it is approached.

Moreover, this gives you a shorcut to compute the expression even quite far away from $x=0$. Try it for $x=\frac \pi {24}$ for which we know the exact values of the trigonometric functions (see [here](http://mathworld.wolfram.com/TrigonometryAnglesPi24.html))
,the exact value of the expression is $-0.4008$ while the above truncated series would give $-0.4023$.
Since $\log(1+x)=x-\frac{x^2}2+O\!\left(x^3\right)$, we have
$$
\begin{align}
&\frac{\log(1+\tan(x))-\sin(x)}{x\sin(x)}\\
&=\frac{\color{#C00}{\tan(x)}\color{#090}{-\frac{\tan^2(x)}2}+\color{#00F}{O\!\left(\tan^3(x)\right)}\color{#C00}{-\tan(x)\cos(x)}}{x\sin(x)}\\
&=\frac{\color{#C00}{\tan(x)}}{x}\frac{\color{#C00}{1-\cos(x)}}{\sin(x)}\color{#090}{-\frac12}\frac{\color{#090}{\tan^2(x)}}{x\sin(x)}+\frac{\color{#00F}{\tan^2(x)}}{x\sin(x)}\color{#00F}{O(\tan(x))}\\
&=\underbrace{\ \frac{\tan(x)}{x}\vphantom{\frac{()}{1+()}}\ }\_1\underbrace{\frac{\sin(x)}{1+\cos(x)}}\_0-\frac12\underbrace{\ \frac{\tan(x)}{x\vphantom{()}}\ }\_1\underbrace{\ \frac1{\cos(x)}\ }\_1+\underbrace{\ \frac{\tan(x)}{x\vphantom{()}}\ }\_1\underbrace{\ \frac1{\cos(x)}\ }\_1\underbrace{O(\tan(x))\vphantom{\frac1{()}}}\_0\\
\end{align}
$$