Find $\lim_{x \to 0} \frac{x + 2\sin(x)}{x+10\cos(x)}$ I am trying to find $\lim_{x \to 0} \frac{x + 2\sin(x)}{x+10\cos(x)}$. 
Using L'Hospital's Rule:
since $\frac{d}{dx}(x + 2\sin(x)) = 1 + [0 \cdot \sin(x)] + [2\cdot\cos(x)]=1 +2\cos(x)$ 
and since  $\frac{d}{dx} (x + 10\cos(x)) = 1 + [0 \cdot \cos(x)] + [10 \cdot -\sin(x)]=1-10\sin(x)$
$$\lim_{x \to 0} \frac{x + 2\sin(x)}{x+10\cos(x)}= \lim_{x \to 0} \frac{1 +2\cos(x)}{1-10\sin(x)}$$
plug in 0 for x:
Since $\cos(0) = 1$, and $\sin(0) = 0$
$$=\frac{1 + 2(1)}{1 - 10(0)} = \frac{1+2}{1-0} = 3$$
But this is apparently wrong.

Why so complicated? The numerator goes to $0$, and the denominator goes to $10$, so the limit is $0$. Also you cannot use L'Hospital's rule here since the limit is not one of the indeterminate forms that permit the use of L'Hospital.
