Hospital rule help

Question

Hospital rule help

anonymous · Answer

$$\lim_{x \rightarrow 0}\frac{ e^{x}-1 }{ \sin(7x) }$$

anonymous · Answer

Well, this goes to 0/0, right?  :)

anonymous · Answer

I think so...

anonymous · Answer

you do need to check that first, otherwise you cannot continue
replace $x$ by $0$ and see exactly what you get

anonymous · Answer

@Dodo1 its "Hopital" not a "Hospital" :)

anonymous · Answer

ops ahah thank you.

anonymous · Answer

@electrokid 
It's actually 
$$L'H\hat opital$$

anonymous · Answer

@PeterPan I know.

anonymous · Answer

what is my first step to solve this

anonymous · Answer

really you have to check that it is $\frac{0}{0}$ first

anonymous · Answer

and btw the "accent circumflex" over the o indicates that historically there was an "s" after the o, so "hospital" is also okay

anonymous · Answer

so first you have to check 
$$\frac{e^0-1}{\sin(7\times 0)}=\frac{0}{0}$$ which is it

anonymous · Answer

then take the derivative sepately top and bottom 
you get 
$$\frac{e^x}{7\cos(7x)}$$ then replace $x$ by $0$

anonymous · Answer

since $\cos(0)=1$ and $e^0=1$ you get 
$$\frac{1}{7}$$

anonymous · Answer

$$L'H\hat opital$$ i can't do it without the italics...

anonymous · Answer

mm wht its 7 cos (7x)?

anonymous · Answer

Thank you satelline