Question

How would I solve lim x -> 0

sin(3x)/x

Answer 1

multiply by 1 in the form of 3/3
then you get

3sin(3x)/(3x) and by a change of variable you get 3sin(u)/u and
\[\lim_{u \rightarrow 0}\frac{ 3\sin u }{ u }=3\lim_{u \rightarrow 0}\frac{ \sin u }{ u }=3\]

Answer 2

Wait, can you eleborate on how you got 3 sinu/u?

Answer 3

let 3x = u then u->0 as x->0.

Answer 4

u=0, right?

Answer 5

you want to get your limit in the form sin(whatever)/whatever because as whatever goes to 0, the limit goes to 1

Answer 6

I know that, but don't you get 3x on the bottom if you multiplied by 3/3?

Answer 7

\[\sin u=u-\frac{1}{3!}u^3+\frac{1}{5!}u^5\pm...\]
\[\sin (3x)=3x-\frac{1}{3!}(3x)^3+\frac{1}{5!}(3x)^5\pm...\]
\[\frac1x\sin (3x)=3-\frac{1}{3!}(3^3x^2)+\frac{1}{5!}(3^5x^4)\pm...\]
at x=0; it all zeroes out but the constant

Answer 8

yes you do\[\frac{ 3 }{ 3 }\frac{ \sin 3x }{ x }=\frac{ 3\sin 3x }{ 3x }\]

Answer 9

What do you do after that?

Answer 10

I got up to that on my own, then I'm stuck.

Answer 11

let 3x = u. as x->0, 3x->0 and u->0. thus \[\lim_{x \rightarrow 0}\frac{ 3\sin 3x }{ 3x }=\lim_{u \rightarrow 0}\frac{ 3\sin u }{ u }=3\lim_{u \rightarrow 0}\frac{ \sin u }{ u }\]

Answer 12

Oh, so you moved both 3's over behind the limit until after all of the simlpification is complete?

Answer 13

Oh, NVM, I get it.

Answer 14

You canceled the 3's from each 3x.

Answer 15

lol, I"m retarded.

Answer 16

no only the top 3. the bottom 3 stays to make 3x in the denominator... same as the argument of the sine function in the numerator

Answer 17

Then you canceled the 3's from the each 3x, right?

Answer 18

Or am I just seeing things?

Answer 19

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