anonymous
  • anonymous
lim x ->0 sin(3x)/(5x^3-4x)
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.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
raffle_snaffle
  • raffle_snaffle
so have you learned L'Hopsitals rule?
anonymous
  • anonymous
I was just getting ready to add that I haven't. Working to get into sinx/x or cosx-1/x
raffle_snaffle
  • raffle_snaffle
okay I am sure we can solve this problem with the lim laws

Looking for something else?

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

More answers

raffle_snaffle
  • raffle_snaffle
|dw:1378769384232:dw|
anonymous
  • anonymous
The issue I've been having is manipulating the functions.
raffle_snaffle
  • raffle_snaffle
I am working on that right now
anonymous
  • anonymous
okay thanks
raffle_snaffle
  • raffle_snaffle
if you take a x^1 and drop it below the exponent changes sign right?
anonymous
  • anonymous
yes negative exponent means it's in the denominator
raffle_snaffle
  • raffle_snaffle
|dw:1378769911488:dw|
raffle_snaffle
  • raffle_snaffle
Boy, I can't figure it out. lol
anonymous
  • anonymous
Yeah me either. Hah
raffle_snaffle
  • raffle_snaffle
I would apply L'Hospitals rule
anonymous
  • anonymous
Yeah, we haven't learned that yet.
raffle_snaffle
  • raffle_snaffle
I wonder if you have to change the sin3x to another trig identity
raffle_snaffle
  • raffle_snaffle
do you have the solution to the question
anonymous
  • anonymous
yeah, it's -3/4
raffle_snaffle
  • raffle_snaffle
you sure it's sin(3x) and not (sin(x))^3
anonymous
  • anonymous
positive i'm looking at it from my book
anonymous
  • anonymous
i think i might know what to do
raffle_snaffle
  • raffle_snaffle
^this guy knows for sure
raffle_snaffle
  • raffle_snaffle
@SithsAndGiggles
anonymous
  • anonymous
\[\lim_{x\to0}\frac{\sin3x}{5x^3-4x}=\lim_{x\to0}\frac{\sin3x}{x\left(5x^2-4\right)}=3\lim_{x\to0}\frac{\sin3x}{3x}\cdot\frac{1}{5x^2-4}\]
anonymous
  • anonymous
yeah i got that far
anonymous
  • anonymous
Right, so the next step would be to split up the limits: \[3\lim_{x\to0}\frac{\sin3x}{3x}\cdot\frac{1}{5x^2-4}=3\left(\lim_{x\to0}\frac{\sin3x}{3x}\right)\left(\lim_{x\to0}\frac{1}{5x^2-4}\right)\]
anonymous
  • anonymous
What do you do with the polynomial in the denominator?
raffle_snaffle
  • raffle_snaffle
Why can you pull a 3 out front? front of the lim?
raffle_snaffle
  • raffle_snaffle
oh nvm Lol
anonymous
  • anonymous
Direct substitution works. @raffle_snaffle, I first multiplied by 3/3, leaving the 1/3 in the denominator and pulling out the 3 from the numerator.
raffle_snaffle
  • raffle_snaffle
I see sithandgigles
anonymous
  • anonymous
Wow, you're right. I've been looking at this too long. There's not an x on 4 after factoring it out.
anonymous
  • anonymous
Thanks.
anonymous
  • anonymous
You're welcome

Looking for something else?

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