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

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. 

$\dfrac{\sin(x^2)}{\sin x} = \dfrac{\sin(x^2)}{x^2} \dfrac{x^2}{x} \dfrac{x}{\sin x}$ tends to 0.
Similarly :
$c\dfrac{\sin(x^2)}{\sin^2 x} = c\dfrac{\sin(x^2)}{x^2} \left( \dfrac{x}{\sin x}\right)^2$ tends to $c$.
Since $1+\cos x$ tends to 2, the result follows.
