Squeeze Theorem Problem to prove a limit..???

Question

they want me to prove that the Limit as "x" approaches 0 for x^3cos(1/x) is equal to zero...i'm kind of confuse can someone give me a detailed explanation on how to prove this using the Squeeze Theorem?

Answer 1

When you have x^3*cos(1/x), you know the max value of cos(1/x) is 1 and the minimum value is -1 (the 1/x function doesn't change the absolute max and min amplitude, only the coefficient in front of the cos does). So the max value of x^3*cos(1/x) is x^3 and the minimum value is -x^3. Taking the lim x->0 for x^3=>0. The lim x->0 for -x^3=>0.

Therefore, since the upper bound is equal to 0 and the lower bound is equal to 0, then since x^3*cos(1/x) is bounded 0 on the top and the bottom, it is equal to 0 by the Squeeze Theorem.

Answer 2

I don't know about the squeeze theorem, but the simple explanation to this problem is that the value of cos never goes beyond +/- 1. Even cos(x) as x->infinity has a defined value between +1 and -1. The x^3 on the other hand, goes to 0 as x ->0. Therefore, you have a situation where as x-> the answer is 0 times some number between 1 and -1. That operation produces 0 no matter what that number is. Hence, the limit is 0.

Answer 3

Squeeze Theorem or sandwich theorem According to which if f(x) is a given function
and g(x) < or = f(x) < or = h(x)
and
g(x) limit x tending to a = L = h(x) limit x tending to a
then f(x) limit x tending to a is also = L

f(x) = x^3.cos(1/x)
as cos varies from -1 to +1 thus g(x) and h(x) can be taken as
x^3*(-1) and x^3*(1) respectively.

-x^3 < = x^3 . cos(1/x) < = x^3

now -x^3 limit x tending to 0 = 0
and x^3 limit x tending to 0 is also 0.
Thus by squeeze theorem x^3.cos(1/x) limit x tending to 0 is also = 0

Answer 4

x^3cos(1/x)=exp{3cos(1/x)lnx}
3cos(1/x)<3 and the lim lnx is equal -infinite so
exp{3cos(1/x)lnx}=0 i.e x^3cos(1/x)=0

Answer 5

when "x" approaches 0
lim f(x) <= lim x^3cos(1/x) <= lim g(x)

First, you will find the functions f(x) and g(x)