Question

Using The Squeeze Theorem to prove that `lim_(x->1)(x-1)^2 sin(5/(x-1) + 3) = 0` ?

Answer 1

Since the sine function's range is `[-1,1]`, and since `lim_{x->1} -(x-1)^2 = lim_{x->1} (x-1)^2 = 0`, the limit exists and is equal to 0.

Explanation:

You cannot do the limit as written as `x->1` using the multiplication limit rule, since the `lim_{x->1} sin (5/(x-1) + 3)` does not exist.

However, based upon the sine function itself, it can be said that the following is always true:

`-1 <= sin (5/(x-1) + 3) <= 1` for all `x != 1`.

Thus, since `(x-1)^2 >= 0` for all `x`, you can multiply all portions of the preceding to arrive at a new relationship:

`-(x-1)^2 <= (x-1)^2 sin (5/(x-1) + 3) <= (x-1)^2`

However, it's evident that the following two limits exist and are equal:

`lim_{x->1} -(x-1)^2 = 0`
`lim_{x->1} (x-1)^2 = 0`

Since each function's limits at `x = 1` exist, and since they are equal, and since the value of the original function is constrained within those two function values, it's limit must also exist and be equal to the same value of 0.

graph{(y - ((x-1)^2)*sin(5/(x-1)+3))(y-(x-1)^2)(y+(x-1)^2) = 0 [-0.021, 2.112, -0.5355, 0.576]}