Compute trigonometric limit without use of de L'Hospital's rule

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

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

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.

With equivalents, it's straightforward: $$\sin u\sim_0 u,\quad 1-\cos x\sim_0 \frac12x^2,\enspace\text{hence}\quad \frac{(x+c)\sin x^2}{1-\cos x}\sim_0\frac{cx^2}{\frac12x^2}=2c.$$