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. 
 A: With equivalents, it's straightforward:
$$\sin u\sim_0 u,\quad 1-\cos x\sim_0 \frac12x^2,\enspace\text{hence}\quad \frac{(x+c)\sin x^2}{1-\cos x}\sim_0\frac{cx^2}{\frac12x^2}=2c.$$
A: 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.
A: $\dfrac{\sin(x^2)}{\sin x} = \dfrac{\sin(x^2)}{x^2} \dfrac{x^2}{x} \dfrac{x}{\sin x}$ 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.
A: Easy to see $\lim_{x\to 0}\frac{ \sin(x^2)}{sin^2 x} = 1$
$$\lim_{x\to 0} \frac{(x+c)\sin(x^2)}{1-\cos(x)} = \lim_{x\to 0} \frac{(x+c) \sin^2 x}{1-\cos(x)} = \lim_{x\to 0} (x+c) (1+\cos(x)) $$
A: $1-\cos x =2\sin^2 \frac {x}{2}= 2(\frac {x}{2})^2 F(x)$ 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. 
A: We can solve in two steps fistly multipling both sides to ${ x }^{ 2 }$ and then to $1+\cos { \left( x \right)  } $ and consider the fact $\lim _{ x\rightarrow 0 }{ \frac { \sin  (x^{ 2 }) }{ x^{ 2 } } =1 } $ we will get

$$\lim _{ x\to 0 } \frac { (x+c)\sin  (x^{ 2 }) }{ 1-\cos  (x) } =\lim _{ x\to 0 } \frac { (x+c){ x }^{ 2 } }{ \left( 1-\cos  (x) \right)  } \frac { \sin  (x^{ 2 }) }{ { x }^{ 2 } } =\\ =\lim _{ x\to 0 } \frac { (x+c){ x }^{ 2 } }{ \left( 1-\cos  (x) \right)  } \frac { \left( 1+\cos  (x) \right)  }{ \left( 1+\cos  (x) \right)  } =\lim _{ x\to 0 } \frac { (x+c){ x }^{ 2 } }{ 1-\cos ^{ 2 }{ \left( x \right)  }  } \left( 1+\cos { \left( x \right)  }  \right) \\ =\lim _{ x\to 0 } \frac { (x+c){ x }^{ 2 } }{ \sin ^{ 2 }{ \left( x \right)  }  } \left( 1+\cos { \left( x \right)  }  \right) =\lim _{ x\to 0 } \left( x+c \right) \left( 1+\cos { \left( x \right)  }  \right) =2c$$

A: 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.
