Question

Limit Question.

Answer 1

\[\lim_{x \rightarrow 0}\frac{ \sin2x }{ \sin3x }\]

Answer 2

2/3

Answer 3

Please Explain :(

Answer 4

ok

Answer 5

one min

Answer 6

since you have indeterminate form, you have to use L'hospital rule

Answer 7

then when you do derivative you get \[\frac{ 2\cos2x }{ 3\cos3x }\]

Answer 8

I dont know l hospitals rule.

Answer 9

you don't?

Answer 10

you know like squeeze theorem?

Answer 11

Please teach me how to solve it,i am helpless :(

Answer 12

oki so there are couple cases when you have indeterminate form, for example 0/0 , inf/inf

Answer 13

so what you do then is basically take the derivative of the top and the bottom

Answer 14

for example \[\lim_{x \rightarrow 1} \frac{ x^3+x^2=2x }{ x-1 }\]

Answer 15

it suppose to be x^3+x^2-2x

Answer 16

so if you plug 1 in you get 0/0 which is indeterminate as it has no meaning. so you have to apply l'hospital, derivative of the top and bottom