Question

lim x->0 sin3x/5x^3-4x.

Answer 1

\[\lim_{x \rightarrow 0}\frac{ \sin 3x }{ 5x ^{2}-4 }=\frac{ \lim_{x \rightarrow 0}3x \frac{ \sin 3x }{ 3x } }{ 5x ^{3}-4x }\]

Answer 2

did u get it

Answer 3

\[=\frac{ 3x \lim_{x \rightarrow 0}\frac{ \sin 3x }{ 3x } }{ 5x ^{3}-4x }\]

Answer 4

u know\[\lim_{x \rightarrow 0}\frac{ \sin 3x }{ 3x }=1\]

Answer 5

now try to do the rest

Answer 6

so would you want to factor out the x?

Answer 7

ya

Answer 8

got it thanks

Answer 9

\[\lim_{x \rightarrow 0}\frac{ 3x }{ 5x ^{3}-4x }\]

Answer 10

welcome

Answer 11

medal????????????????

Answer 12

i dont know how to do that or what means.. I am sorry I just found this site

Answer 13

just click on best answer

Answer 14

Hey did you by any chance see my other question?

Answer 15

ya