anonymous
  • anonymous
2nd question: Find lim (sin2xcot4x) X->0.
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.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
are you allowed to use l'hopital's rule?
anonymous
  • anonymous
assuming this is \[\lim_{x\rightarrow 0}\sin(2x)\cot(4x)\] \[\lim_{x\rightarrow 0}\frac{\sin(2x)\cos(4x)}{\sin(4x)}\]
nenadmatematika
  • nenadmatematika
1/2

Looking for something else?

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

More answers

anonymous
  • anonymous
in any case the answer is \[\frac{1}{2}\]
anonymous
  • anonymous
but if you need to show your work your answer will depend on what you are allowed to use. l'hopital's rule is the simplest, otherwise it will be some work
anonymous
  • anonymous
i dont know the hospital rule that you have said :(
anonymous
  • anonymous
you mean you do not know it or you are not allowed to use it? i assume this is calc class, so hve you covered deriviatives yet or are you just starting out?
nenadmatematika
  • nenadmatematika
|dw:1328531016861:dw|
nenadmatematika
  • nenadmatematika
without using l'hopital rule
anonymous
  • anonymous
any computation as a long the answer will be the same in a long method
anonymous
  • anonymous
i dont know it sorry :(
anonymous
  • anonymous
there is a nice neat answer above. i think you can also use \[\frac{sin(2x)\cos(4x)}{\sin(4x)}=\frac{\sin(2x)(\cos^2(2x)-\sin^2(2x))}{2\sin(2x)\cos(2x)}\]
anonymous
  • anonymous
here i used the double angle formula for sine and cosine
anonymous
  • anonymous
can i post the 3rd question too?
anonymous
  • anonymous
then cancel the sin(2x) top and bottom, get \[\frac{\cos^2(2x)-\sin^2(2x)}{2\cos(2x)}\] replace x by zero, get \[\frac{1}{2}\]
anonymous
  • anonymous
no limit to the amount you can post
anonymous
  • anonymous
yes i will write your solution :)
anonymous
  • anonymous
if f(x)=tanx-x and g(x)=x^3, evaluate the limit of f(x) over g(x) as x approaches 0. -3rd question.
anonymous
  • anonymous
@nenadmatematica if you can do this without l'hopital or power series i will be impressed
nenadmatematika
  • nenadmatematika
haha I just wanted to ask you the same thing :D ....
anonymous
  • anonymous
lol well, i guess we cannot give an elementary reason for this. the answer is \[\frac{1}{3}\] but i cannot think of a gimmick to simplify this expression. are you sure you have not covered l'hopital? because i am stumped. in particular you have a trig fuction combined in combination with x and x^3 so there is no simple trig identity that will change the form of this for you
anonymous
  • anonymous
@nenadmatematika tahnks for helping us too :)
nenadmatematika
  • nenadmatematika
well you're welcome....I agree with satellite that this example is very convenient for using L'Hopital rule.....I can't think of any other way now :D
anonymous
  • anonymous
i really dont know but if both of you wants to use L'hopital rule. then i will agree to both of you
anonymous
  • anonymous
4th question: Evaluate the lim x^2-16 over x+4 as x->4.

Looking for something else?

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