# Discontinuous/Continuous Functions

1. Sep 27, 2014

### happysmiles36

1. The problem statement, all variables and given/known data

Is the function continuous at x=1?

f(x) =
When x is less than 1:
$$(x-1)^2sin(\frac{1}{x-1})$$
When x is greater or equal to 1:
$$xsin(πx)$$

2. Relevant equations
None

3. The attempt at a solution
The answer is yes, don't know how to show that, this isn't homework just studying for exams >.<.

$$\lim_{x\rightarrow 1+} {xsin(πx)} = 0$$
----------------------------------
$$\lim_{x\rightarrow 1-} {(x-1)^2sin(\frac{1}{x-1})}$$

Here I used:
$$\lim_{x\rightarrow 0} {\frac{sin(x)}{x}} = 1$$
But I don't think you can do this, so I don't know how else to solve this.

$$\lim_{x\rightarrow 1-} {(x-1)} * \lim_{x\rightarrow 0} {\frac{sin(\frac{1}{x-1})}{\frac {1}{x-1}}} = 0 * 1 = 0$$

Help explaining how to do this question would be helpful.:)

2. Sep 27, 2014

### jz92wjaz

You can't do that, but there is a really simple trick. What are the maximum and minimum values you can have for sin(x)? Are they finite or infinite?

3. Sep 27, 2014

### happysmiles36

Sin(x) goes between 1 and -1 and it oscillates to infinity and negative infinity.

I ended up thinking that as the limit approaches 1 from the left side,
(x-1)^2 becomes 0 and sin(1/(x-1)) becomes sin(-infinity), so it will be 0 * a number between 1 and -1, and that results in a 0. Is this valid? because sin (-infinity) is undefined, so I don't know if this would be correct or is there a way to show this without the sin being undefined.

Last edited: Sep 27, 2014
4. Sep 27, 2014

### jz92wjaz

It's correct. When dealing with limits, a finite number multiplied by zero is still zero, even if you don't know exactly what the finite number is.

5. Sep 27, 2014

Thanks :)