anonymous
  • anonymous
LIMITS ALSO!
Calculus1
  • Stacey Warren - Expert brainly.com
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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!
anonymous
  • anonymous
?
Jhannybean
  • Jhannybean
Well limits are fun.
anonymous
  • anonymous
The limit of a question mark. idk tbh.

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anonymous
  • anonymous
whats the question by chance?
anonymous
  • anonymous
someone give me a medal for that hilarious joke?
primeralph
  • primeralph
Hahahahahahahahahahahahahahahahhahaha Not funny.
anonymous
  • anonymous
oh sush
anonymous
  • anonymous
@PatrickJordon ty. its nice to see that someone other than me has a good sense of humour.
Jhannybean
  • Jhannybean
What is the question mark approaching? and from which side?
primeralph
  • primeralph
@genius12 You know I'm kidding
anonymous
  • anonymous
:D np
anonymous
  • anonymous
it's an infinitely large question mark @Jhannybean
Jhannybean
  • Jhannybean
:P
anonymous
  • anonymous
|dw:1371411495511:dw|
anonymous
  • anonymous
it's the a semicircle for 1 < x < 3 and x^3 for 1.5 < x < 3. its not a function. and there is a circle of radius 1 at the bottom. you must find the limit as x --> 0
anonymous
  • anonymous
what's between sin(x) and sin(2x)? @markragay
Jhannybean
  • Jhannybean
sin (2x)= 2sin(x)cos(x)
anonymous
  • anonymous
none...
anonymous
  • anonymous
are they being multiplied? @markragay
anonymous
  • anonymous
yes
primeralph
  • primeralph
|dw:1371357618688:dw|
anonymous
  • anonymous
is this one of those annoying "i can't use l'hopital" questions, or is it allowed?
primeralph
  • primeralph
|dw:1371357741805:dw|
Jhannybean
  • Jhannybean
lol @satellite73
primeralph
  • primeralph
|dw:1371357780950:dw|
Jhannybean
  • Jhannybean
Good job.
anonymous
  • anonymous
thank you .....
primeralph
  • primeralph
You're welcome.
anonymous
  • anonymous
what is o divided by zero? is it infinity?
anonymous
  • anonymous
Use l'hopitals rule and differentiate the top and bottom twice:\[\bf \lim_{x \rightarrow 0}\frac{ \sin(x)\sin(2x) }{ 1-\cos(x)}=\lim_{x \rightarrow 0}\frac{ \cos(x)\sin(2x)+2\cos(2x)\sin(x) }{ \sin(x) }\]\[\bf =\lim_{x \rightarrow 0}\frac{ -\sin(x)\sin(2x)+2\cos(2x)\cos(x)-4\sin(2x)\sin(x)+2\cos(x)\cos(2x) }{ \cos(x) }\]\[=4\]
anonymous
  • anonymous
@markragay
primeralph
  • primeralph
@genius12 Right on time.
anonymous
  • anonymous
zero ddivided by zero is it infinity?
anonymous
  • anonymous
@primeralph only an extra thing to know. i already knew that the question had been answered already.
anonymous
  • anonymous
@markragay it's indeterminate.
primeralph
  • primeralph
@genius12 Thanks for making it clearer.

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