Trouble with l'Hôpital's rule for $\lim_{x\rightarrow 0} \frac{4x+4\sin x}{10x+10\cos x}$ This is an assignment and I am stuck:

Find the limit, whether finite or infinite, or indicate that the limit does not exist. Use l'Hôpital's Rule if appropriate.
$$\lim_{x\rightarrow 0} \frac{4x+4\sin x}{10x+10\cos x}.$$

When I try l'Hôpital's rule I get $\frac 8{10}$. Can someone tell me why this isn't right?
 A: Original posted question:
$$\lim_{x\rightarrow 0} \frac{4x+4\sin x}{10x+10\sin x} = \lim_{x\to 0} \frac{(4x + 4\sin x)'}{(10x + 10 \sin x)'} = \lim_{x\to 0} \frac{4+4\cos x}{10 + 10\cos x} = \frac {4 + 4}{10 + 10} = \frac {8}{20} = \frac 25$$

Since you meant to post $$\lim_{x\rightarrow 0} \frac{4x+4\sin x}{10x+10\cos x}$$ note that in this case, l'Hôpital is not applicable (the limit does not at first evaluate to an indeterminate limit). Nor would we want to use it! It is easily solved by evaluating immediately: 
$$\lim_{x\rightarrow 0} \frac{4x+4\sin x}{10x+10\cos x} = \frac{ 0 + 0}{0 + 10(1)} = \frac {0}{10} = 0$$
IMPORTANT TO REMEMBER: We apply l'Hôpital's rule if and only if a limit evaluates to an indeterminate form. That bold-face link will take you to Wikipedia's concise list of "what counts" as an indeterminate form, and why.
A: Just plug in! The denominator is $10$, not zero. ;)
A: It may be a non-rigorous approach but here you go :
When $x\to 0$ we have these two approximations :
$$\sin x \overset{x\to 0}\sim x \ \ \ \ \ \text{and}\ \ \ \ \  \cos(x) \overset{x\to 0}\sim 1-\frac{x^2}{2}$$
Let :
$$\mathcal{L} =\lim_{x\rightarrow 0} \frac{4x+4\sin x}{10x+10\cos x}$$
Using those approximations we get :
\begin{align}
\mathcal{L}&=\lim_{x\to 0} \frac{4x+4x}{10x-10-5x^2}\\
&=\lim_{x\to 0} \frac{8x}{-5x^2+10x-10}\\
&= \frac{0}{10}\\
&=0
\end{align}
Your limit doesn't need anything it's already equals $0$ without direct substitutions.
