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. 

Your first step is very good: by multiplying numerator and denominator by $1+\cos x$, the limit becomes
$$
\lim_{x\to0}(x+c)(1+\cos x)\frac{\sin(x^2)}{\sin^2 x}=
\lim_{x\to0}(x+c)(1+\cos x)\frac{\sin(x^2)}{x^2}\frac{x^2}{\sin^2 x}
$$
and it should be easy to finish.
