Question

Using l'Hopitals rule to find
Lim x-->0 sin5x/sinx, I will post my attempt inside would like someone to check

Answer 1

\[\lim_{x \rightarrow 0} \frac{ \sin5x }{ sinx } == 5\]

Answer 2

so first i took \[\lim_{x \rightarrow 0} \frac{ \sin5x }{ sinx }\] and then took the \[\frac{ d }{ dx }\] of top and bottom and got

\[\lim_{x \rightarrow 0} \frac{ 5\cos5x }{ cosx }\] then evaluated the equation using the limit

Answer 3

looks good

Answer 4

it came to 5 *(cos(infinity)) which is 5*1 which is 5

and cos(infinity) = 1

so 5/1 = 5

Answer 5

Thank you

Answer 6

yep
you only needed one application of lhopitals tis time.

Answer 7

awesome thank you again

Answer 8

yw