# Another math test question

1. Oct 7, 2011

### davie08

1. The problem statement, all variables and given/known data
i think I remember the question to be this:

lim (y-3)/(y-3)(y+3)
y→3

it was the first limit question and the easiest but im not sure if I did it right.

2. Relevant equations

3. The attempt at a solution

so what I did was I cancelled the y-3's out to get:

1/(y+3) and this is what I was unsure about is that the first step and after this I got:

1/(3+3) = 1/6

2. Oct 7, 2011

### DivisionByZro

You're correct:

$$\lim_{y \to \ 3} \frac{y-3}{(y-3)(y+3)} = \lim_{y \to \ 3 } \frac{1}{y+3} = \frac{1}{6}$$

3. Oct 7, 2011

### spamiam

You have the correct answer, but it's probably a good idea to try to justify your last step. When is it true that $\lim_{y \to a} f(y) = f(a)$? What can you say about the continuity of rational functions?

4. Oct 7, 2011

### HallsofIvy

A crucial, and often overlooked property of limits is this: if f(x)= g(x) for all x except x= a, then $\limit_{x\to a} f(x)= \limit_{x\to a}g(x)$.

For this problem, you can correctly say that
$$\frac{y-3}{(y-3)(y+3)}= \frac{1}{y+3}$$
for all x except x= 3 and so the limits are the same.

Actually, "what can you say about the continuity of rational functions?" is NOT a good question here because continuity also involves the value of the function as well as the limit. 1/(x+3) is continuous at x= 3, so its limit is 1/(3+3)= 1/6 while (x-3)/(x-3)(x+3) is NOT continuous at x= 3.

5. Oct 7, 2011

### spamiam

This statement is true exactly because rational functions are continuous anywhere their denominator isn't zero, which is why I asked the question. I guess I should have mentioned the fact that there is a removable singularity at x=3, but the OP seemed to handle it appropriately.