<s>[INST] 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. 
 [/INST] Using Taylor series make life "easy". Start with $$\sin(x)=x-\frac{x^3}{6}+O\left(x^4\right)$$ $$\sin(x^2)=x^2-\frac{x^6}{6}+O\left(x^8\right)$$ $$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}+O\left(x^6\right)$$ So,$$\frac{(x+c)\sin(x^2)}{1-\cos(x)}=\frac{(x+c)\left(x^2-\frac{x^6}{6}+O\left(x^8\right) \right)}{\frac{x^2}{2}-\frac{x^4}{24}+O\left(x^6\right)}$$ Now, long division to get $$\frac{(x+c)\sin(x^2)}{1-\cos(x)}=2 c+2 x+\frac{c x^2}{6}+\frac{x^3}{6}+O\left(x^4\right)$$ which shows the limit and how it is approached.
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