$\lim_{x\to 0}\frac{e^x-1}{\sin x}$ equal to $\lim_{x\to 0}\frac{e^x-1}{x}$ because $x$ and $\sin x$ tend both to $0$ for ${x\to 0}$ I'm stuck in this limit:
$$\lim_{x\to 0}\frac{x(e^x-1)}{\cos x-1}$$
I tried to solve it using special limits, so:
$$\lim_{x\to 0}\frac{x(e^x-1)}{\cos x-1}=$$
$$=\lim_{x\to 0}(e^x-1)\frac{x(\cos x+1)}{(\cos x-1)(\cos x+1)}=$$
$$=\lim_{x\to 0}-\frac{x}{\sin^2 x}(e^x-1)(\cos x+1)=$$
$$=\lim_{x\to 0}-\frac{x}{\sin x}\cdot\frac{e^x-1}{\sin x}\cdot(\cos x+1)$$
I know that $\lim_{x\to 0}\left(\frac{x}{\sin x}\right)=1$
I know also the special limit $\lim_{x\to 0}\left(\frac{e^x-1}{x}\right)=1$
My question is: can I intend $\lim_{x\to 0}\left(\frac{e^x-1}{\sin x}\right)$ equal to $\lim_{x\to 0}\left(\frac{e^x-1}{x}\right)$ because $x$ and $\sin x$ tend both to $0$ for ${x\to 0}$?
If not, how can I solve my limit, without series or something difficult?
Thank you!
 A: Yes, you can says that $\lim_{x\to 0}\dfrac{e^x-1}{\sin x}$ equal to $\lim_{x\to 0}\dfrac{e^x-1}{x}$, but not just because $x$ and $\sin x$ tend both to $0$ for ${x\to 0}$, this is not enough.
You can write
$$\lim_{x\to 0}\dfrac{e^x-1}{\sin x}=\lim_{x\to 0}\dfrac{e^x-1}{\sin x}\frac xx=\lim_{x\to 0}\dfrac{e^x-1}x\frac x{\sin x}=\lim_{x\to 0}\dfrac{e^x-1}x\lim_{x\to 0}\frac x{\sin x}.$$
Now you get the equivalence from
$$\lim_{x\to 0}\frac x{\sin x}=1.$$
This is justified by the property that the limit of a product is the product of the limits, provided they exist.
A: $$\lim _{ x\rightarrow 0 }{ \frac { x\left( { e }^{ x }-1 \right)  }{ \cos { x } -1 } = } \lim _{ x\rightarrow 0 }{ \frac { { x }^{ 2 }\left( { e }^{ x }-1 \right)  }{ x\left( \cos { x } -1 \right)  } = } \lim _{ x\rightarrow 0 }{ \frac { \left( { e }^{ x }-1 \right)  }{ x } \lim _{ x\rightarrow 0 }{ \frac { { x }^{ 2 } }{ -2\sin ^{ 2 }{ \frac { x }{ 2 }  }  } = }  } \\ =\lim _{ x\rightarrow 0 }{ \frac { \left( { e }^{ x }-1 \right)  }{ x } \lim _{ x\rightarrow 0 }{ \frac { { x }^{ 2 } }{ -2\sin ^{ 2 }{ \frac { x }{ 2 }  }  } = }  } -\frac { 1 }{ 2 } \lim _{ x\rightarrow 0 }{ \frac { \left( { e }^{ x }-1 \right)  }{ x } \lim _{ x\rightarrow 0 }{ \frac { 4 }{ { \left( \frac { \sin { \frac { x }{ 2 }  }  }{ \frac { x }{ 2 }  }  \right)  }^{ 2 } } = }  } -2$$
A: If $\lim_{x\to0}f(x)=0$  and $\lim_{x\to0}g(x)=0$ than $\lim_{x\to0}\frac{f(x)}{g(x)}=\lim_{x\to0}\frac{f'(x)}{g'(x)}$. 
This is the L'Hospital rule. 
Use that for your limit:
$$\lim_{x\to0}\frac{x(e^x-1)}{\cos x-1}$$
$$f(x)=x(e^x-1)$$
$$f'(x)=(x+1)e^x-1$$
$$g(x)=\cos x-1$$
$$g'(x)=-\sin x$$
$$\lim_{x\to0}\frac{f(x)}{g(x)}=\lim_{x\to0}\frac{f'(x)}{g'(x)}$$
$$\lim_{x\to0}\frac{x(e^x-1)}{\cos x-1}=\lim_{x\to0}\frac{(x+1)e^x-1}{-\sin x}$$
Use the rule again:
$$f''(x)=(x+2)e^x$$
$$g''(x)=-\cos x$$
$$\lim_{x\to0}\frac{x(e^x-1)}{\cos x-1}=\lim_{x\to0}\frac{(x+1)e^x-1}{-\sin x}=\lim_{x\to0}\frac{(x+2)e^x}{-\cos x}$$
$$\lim_{x\to0}\frac{(x+2)e^x}{-\cos x}=\frac{(0+2)e^0}{-\cos 0}=-2$$
