Question

can someone explain how to solve lim x approaching 0 cos(2x)-1/sinx

Answer 1

what math class are you in?

Answer 2

\[\lim_{x \rightarrow 0}\frac{ \cos(2x) - 1}{ \sin(x) }\]

Answer 3

what happens to the numerator when x->0

Answer 4

sinx would be 0

Answer 5

just the numerator

Answer 6

top part of the fraction

Answer 7

0

Answer 8

and the denominator is going to 0 as well like you said

Answer 9

so its going to a form of \[\frac{ 0 }{ 0 }\]

Answer 10

This is called an 'Indeterminate Form" I am assuming you are in a calculus class?

Answer 11

yea

Answer 12

and have you learned what a derivative is yet?

Answer 13

yea

Answer 14

there is a rule called " L'Hopital's Rule " it says that when a limit takes an indeterminate form: then \[\lim_{x \rightarrow 0} \frac{ \cos(2x)-1 }{ \sin(x) } = \lim_{x \rightarrow 0}\frac{ \frac{ d }{ dx }(\cos(2x)+1) }{ \frac{ d }{ dx }(\sin(x)) }\]

Answer 15

where d/dx means take the derivative.

Answer 16

Can you take the derivative of both of those?

Answer 17

ok so it would be (-2sin2x)/(cosx) , plugging in 0 would equal 0/1=0

Answer 18

Absolutely correct!