show limit x to 1 (x^2)sin(1/(x-1)) = 0
To show that the limit of the given function as x approaches 1 is 0, we need to use the definition of the limit and apply some algebraic simplifications.
Let's break it down step by step:
1. Start with the given function:
lim(x → 1) (x^2)sin(1/(x - 1))
2. Notice that the term sin(1/(x - 1)) is undefined at x = 1, so we can't simply substitute x = 1 into the function. However, we can make use of the fact that the sine function is bounded between -1 and 1:
-1 ≤ sin(1/(x - 1)) ≤ 1
3. Multiplying each side of the inequality by (x^2), we get:
-x^2 ≤ (x^2)sin(1/(x - 1)) ≤ x^2
4. Now, we can apply the limit law of Squeeze Theorem. As x approaches 1, the bounds of -x^2 and x^2 both approach 0:
lim(x → 1) -x^2 ≤ lim(x → 1) (x^2)sin(1/(x - 1)) ≤ lim(x → 1) x^2
0 ≤ lim(x → 1) (x^2)sin(1/(x - 1)) ≤ 0
5. Since the lower bound and the upper bound of the function both approach 0, we can conclude that the limit of (x^2)sin(1/(x - 1)) as x approaches 1 is indeed 0:
lim(x → 1) (x^2)sin(1/(x - 1)) = 0
This shows that the given function has a limit of 0 as x approaches 1.