Solve Limiting Value with Squeeze Theorem: x-1

[candynrg]

How do I solve:
lim [(x^2-2x+1)cos(1/x^2-1)]=0
x-1

 

[dextercioby]

That limit is zero.

\left(x^{2}-2x+1\right)(-1)\leq \left(x^{2}-2x+1\right)\left(\cos\frac{1}{x^{2}-1}\right)\leq \left(x^{2}-2x+1\right)(+1)

Take limit in all 3 of them and u'll have your answer.

Daniel.

 

[M98Ranger]

To solve this limit using the Squeeze Theorem, we need to find two functions that are both greater than or equal to the given function and whose limits as x approaches 1 are equal to 0.

One possible choice for the upper bound function is f(x) = x^2 - 2x + 1, since it is always greater than or equal to the given function and its limit as x approaches 1 is equal to 0.

For the lower bound function, we can use g(x) = -x^2 + 2x - 1, which is always less than or equal to the given function and also has a limit of 0 as x approaches 1.

Therefore, we have:

-g(x) ≤ x^2 - 2x + 1 ≤ f(x)

Taking the limit as x approaches 1 for all three functions, we get:

-lim g(x) = 0 ≤ lim (x^2 - 2x + 1) ≤ lim f(x) = 0

By the Squeeze Theorem, since the upper and lower bound functions have the same limit as x approaches 1, the given function must also have a limit of 0.

Therefore, the solution to the limit is 0.