Aaron R.

asked • 02/04/18

Find the limit as x approaches 0 for the function g(x) given -1<= | ((x^2)g(x) / (1-cosx)^2)| <= 1 using Squeeze Theorem

Hello! I was wondering how I could solve this problem. I'm not quite sure what are needed to be put on both left and right side of the equation ( -1 <= and <= 1).

1 Expert Answer

By:

Alexandr K. answered • 02/04/18

Tutor
5 (7)

The life of a Math teacher is to give you problems.

See tutors like this
See tutors like this
Multiply through by (1 - cosx)^2
You can ignore the absolute value when doing this because anything squared is always positive.
Then apply the squeeze theorem.

If you have any particular questions let me know.
Upvote • 0 Downvote
More
Report

Aaron R.

May I ask what you mean about "multiply through?" 
 
will it turn out to be 
 
-x^2 / (1-cosx)^2 <= ((x^2)(g(x)) / (1-cosx)^2) <= x^2 / (1-cosx)^2 ? 
 
Thank you
Report

02/05/18

Alexandr K.

Treat the boundary like any equation, what you do to one part, do the same to all the others, including the middle part.
 
-1 <= ((x^2)g(x)/(1 - cosx)^2) <= 1
 
Multiply each part by (1 - cosx)^2
 
-(1 - cosx)^2 <= (x^2)g(x) <= (1 - cosx)^2
 
Just curious, have you done L'Hopital's rule yet?
Report

02/05/18

Aaron R.

Oh. Then I'll divide both sides with x^2?
 
They're asking us to answer this without L'Hopital's yet, and nope we have not discussed that lesson.
Report

02/05/18

Alexandr K.

If you were using L'Hopital's, then yes you would just divide through by x^2 and then apply the squeeze theorem. But the limits at the ends would end up being in the indeterminant form 0/0.
 
Was the function g(x) defined at all in this problem?
Report

02/05/18

Alexandr K.

Okay, this is a bit of a tricky problem, it almost stumped me for a second.
First, we will multiply through to get |g(x)| in between two other functions.
 
We get -((1 - cosx)^2)/x^2 <= |g(x)| <= ((1 - cosx)^2)/x^2
 
Then we apply squeeze theorem by taking the limits as x approaches 0 of the ends.
 
lim ((1 - cosx)^2)/x^2 as x -> 0
 
This is a tricky limit without L'Hopital's Rule. I took some time and solved it, give it a shot. Hint: You have to use the multiplication property of limits and you have to use the squeeze theorem again on one of the new limits.
 
Let me know if you still can't figure it out.
Report

02/05/18

Aaron R.

Funny though cause that's all they mentioned - the one I typed in the header. No definition for g(x) at all. 
Report

02/05/18

Alexandr K.

Granted this is not an easy problem, but it's certainly do-able. Read my above comment and give it ananoth shot. Let me know if you get completely lost  
Report

02/05/18

Aaron R.

so I got a limit of -1 for the left side I'm not quite sure if what I did is allowed. 
 
First, I squared the (1 - cosx) so I got - (1-2cosx+ cos^2(X)) / x^2
 
Then, changed 1 -cos^2(X) to sin^2(x) , so -sin^2(X) -2cosx / x^2
 
then manipulated it in a way that I chopped them into pieces so I got 
 
(- 1/1) ( ( sinx/x ) ( sin x/x) ) - ( ( 2/1) (cosx / x) ( 1 / x)) 
 
so it ends up as (-1/1) ( (1)(1))  - ( (2) (0) (1/x) ) =  -1(1-0) = -1 
Report

02/05/18

Aaron R.

positive 1 rather
Report

02/05/18

Alexandr K.

Very close, but you made a mistake converting trig identities. You have 1 + Cos^2(x), not 1 minus.
 
You're on the right track though, instead of multiplying out the square, try taking out the squares outside the limit, then multiply the inside of the limit by the conjugate of 1 - cosx. See where that takes you. Give it another shot.
Report

02/05/18

Aaron R.

I got 0 as the limit 
Report

02/05/18

Alexandr K.

You got it. So by the squeeze theorem, the limit of g(x) should be zero as well. Good job! 
Report

02/05/18

Aaron R.

It was fun. Thanks! I finally got the procedure in answering these types of problems. :)
Report

02/05/18

Still looking for help? Get the right answer, fast.

Ask a question for free

Get a free answer to a quick problem.
Most questions answered within 4 hours.

OR

Find an Online Tutor Now

Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.