anonymous
  • anonymous
How would I solve lim x -> 0 sin(3x)/x
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
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 \[\lim_{u \rightarrow 0}\frac{ 3\sin u }{ u }=3\lim_{u \rightarrow 0}\frac{ \sin u }{ u }=3\]
anonymous
  • anonymous
Wait, can you eleborate on how you got 3 sinu/u?
anonymous
  • anonymous
let 3x = u then u->0 as x->0.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
u=0, right?
anonymous
  • anonymous
you want to get your limit in the form sin(whatever)/whatever because as whatever goes to 0, the limit goes to 1
anonymous
  • anonymous
I know that, but don't you get 3x on the bottom if you multiplied by 3/3?
amistre64
  • amistre64
\[\sin u=u-\frac{1}{3!}u^3+\frac{1}{5!}u^5\pm...\] \[\sin (3x)=3x-\frac{1}{3!}(3x)^3+\frac{1}{5!}(3x)^5\pm...\] \[\frac1x\sin (3x)=3-\frac{1}{3!}(3^3x^2)+\frac{1}{5!}(3^5x^4)\pm...\] at x=0; it all zeroes out but the constant
anonymous
  • anonymous
yes you do\[\frac{ 3 }{ 3 }\frac{ \sin 3x }{ x }=\frac{ 3\sin 3x }{ 3x }\]
anonymous
  • anonymous
What do you do after that?
anonymous
  • anonymous
I got up to that on my own, then I'm stuck.
anonymous
  • anonymous
let 3x = u. as x->0, 3x->0 and u->0. thus \[\lim_{x \rightarrow 0}\frac{ 3\sin 3x }{ 3x }=\lim_{u \rightarrow 0}\frac{ 3\sin u }{ u }=3\lim_{u \rightarrow 0}\frac{ \sin u }{ u }\]
anonymous
  • anonymous
Oh, so you moved both 3's over behind the limit until after all of the simlpification is complete?
anonymous
  • anonymous
Oh, NVM, I get it.
anonymous
  • anonymous
You canceled the 3's from each 3x.
anonymous
  • anonymous
lol, I"m retarded.
anonymous
  • anonymous
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
anonymous
  • anonymous
Then you canceled the 3's from the each 3x, right?
anonymous
  • anonymous
Or am I just seeing things?
anonymous
  • anonymous
\[\lim_{x \rightarrow 0}\frac{ 3\sin 3x }{ 3x }=3\lim_{x \rightarrow 0}\frac{ \sin 3x }{ 3x }=3\lim_{u \rightarrow 0}\frac{ \sin u }{ u }\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.