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 $-0.4023$.
You could expand the transcendental functions in the limitand as their Maclaurin series up to second order: since in a neighborhood of $0$ we have
$$\begin{split}
\tan x = x + x^2 p(x), \quad \sin x = x + x^2 q(x), \quad \ln(x+1) = x- \frac 1 2 x^2 + x^2 r(x)
\end{split}$$
for some functions $p,q,r$ that vanish as $x \to 0$, then
$$\begin{split}
L &=
\lim\_{x\to 0} \frac{\ln(x + x^2p(x) + 1)-x - x^2q(x)}{x(x + x^2q(x))} \\
&= \lim\_{x\to 0} \frac{1}{1+x q(x)}\frac{(x + x^2 p(x)) - \frac 1 2 (x + x^2p(x))^2 + (x+x^2p(x))^2 r(x)-x- x^2q(x)}{x^2} \\
&= 1 \cdot \lim\_{x\to 0}\left( -\frac{1}{2} + p(x)(1-x-x^2p(x)/2)+r(x)(1+2xp(x)+x^2p(x)^2)-q(x) \right)= -\frac 1 2.
\end{split}$$
There is a short-hand notation that can help simplifying the above clutter: one writes
$$f(x) = g(x) + \mathtt o(h(x)) \quad \text{as}\ x\to 0 $$
if there exists a function $s$ defined in a neighborhood of $0$ such that $f(x) = g(x) + h(x) s(x)$ in that neighborhood. Thus $\tan x = x + \mathtt o(x^2)$, $\sin x = x + \mathtt o(x^2)$ and $\ln(x+1) = x - \frac 1 2 x^2 + \mathtt o(x^2)$. Take a look at [this wiki page](https://en.wikipedia.org/wiki/Big_O_notation#Little-o_notation) for further information.