TITLE: Compute trigonometric limit without use of de L'Hospital's rule
QUESTION [8 upvotes]: $$ \lim_{x\to 0} \frac{(x+c)\sin(x^2)}{1-\cos(x)}, c \in \mathbb{R^+} $$

Using de L'Hospital's rule twice it is possible to show that this limit equals $2c$. However, without the use of de L'Hospital's rule I'm lost with the trigonometric identities.
I can begin by showing 
$$
\lim\frac{x\sin (x^2)(1+\cos(x))}{\sin^2x}+\frac{c\sin x^2(1+\cos x)}{\sin^2x}=\lim\frac{\sin (x^2)(1+\cos(x))}{\sin x}+\frac{c\sin x^2(1+\cos x)}{\sin^2x},
$$ 
and here I'm getting stuck. I will appreciate any help.

REPLY [2 votes]: $1-\cos x =2\sin^2 \frac {x}{2}= 2(\frac {x}{2})^2 F(x)$ where $\lim_{x\to 0}F(x)=1.$
$\sin x^2=x^2G(x)$ where  $\lim_{x\to 0}G(x)=1.$
$\frac {(x+c)\sin x^2}{1-\cos x}=\frac {(x+c)x^2 G(x)}{2(\frac {x}{2})^2F(x)}=(2x+2c)\frac {F(x)}{G(x)}.$ 
The rest is obvious.