Question

Hospital rule help

Answer 1

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

Answer 2

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

Answer 3

I think so...

Answer 4

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

Answer 5

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

Answer 6

ops ahah thank you.

Answer 7

@electrokid
It's actually
\[L'H\hat opital\]

Answer 8

@PeterPan I know.

Answer 9

what is my first step to solve this

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

mm wht its 7 cos (7x)?

Answer 17

Thank you satelline