Question

How do you use the Squeeze Theorem to find `lim(x-1)sin(pi/x-1)` as x approaches one?

Answer 1

`lim_(x rarr 1) (x-1)sin(pi/x - 1) = 0`

Explanation:

Firstly, we don't need the squeeze theorem because the function is continuous at values close to `x=1`, so the limit is `f(1)`, that being said, to use the squeeze theorem we must remember that

`-1 <= sin(theta) <= 1`

If we have `theta = pi/x - 1`, we have

`-1 <= sin(pi/x - 1) <= 1`

Multiplying both sides by `x-1` we have

`1 -x <= (x-1)sin(pi/x - 1) <= x - 1` for `x >0`

Since

`lim_(x rarr 1) 1 - x = lim_(x rarr 1) x - 1 = 0`

The squeeze theorem tells us that

`lim_(x rarr 1) (x-1)sin(pi/x - 1) = 0`

Answer 2

`lim_(xrarr1)(x-1)sin(pi/(x-1)) = 0`

Explanation:

`-1 <= sin(pi/(x-1)) <= 1` for all `x != 1`

Limit from the right is 0

For `x > 1`, we have `x-1 > 0` so we can multiply the inequality without changing the inequalities:

`-(x-1) <= (x-1)sin(pi/(x-1)) <= x-1` for all `x != 1`

`lim_(xrarr1^+) -(x-1) = 0` and `lim_(xrarr1^+) (x-1) = 0`

So `lim_(xrarr1^+) (x-1)sin(pi/(x-1)) = 0`

Limit from the left is 0

For `x < 1`, we have `x-1 < 0` so when we multiply the inequality we must change the inequalities:

`-(x-1) >= (x-1)sin(pi/(x-1)) >= x-1` for all `x != 1`

`lim_(xrarr1^-) -(x-1) = 0` and `lim_(xrarr1^-) (x-1) = 0`

So `lim_(xrarr1^-) (x-1)sin(pi/(x-1)) = 0`

Therefore,

`lim_(xrarr1) (x-1)sin(pi/(x-1)) = 0`