# Limit of

1. Sep 26, 2004

### buffgilville

1) Compute the limit of f(x) = -(x-3)/square root of (x^2 - 9)
as x approaches 3 from the right.

2) Compute the limit of f(x) = - (x-3)/absolute value of (x-3)
as x approaches 3 from the left.

I got does not exist for both. Is that right?

Last edited: Sep 26, 2004
2. Sep 26, 2004

### arildno

No that is incorrect.
Let x>3
Then
$$\sqrt{x^{2}-9}=\sqrt{x-3}\sqrt{x+3}$$
Or:
$$-\frac{x-3}{\sqrt{x^{2}-9}}=-\sqrt{\frac{x-3}{x+3}}$$
Hence, the right-hand side limit when x goes to 3 is 0.
(The limit from the left-hand side doesn't make much sense for real f, since there are no x<3 which yields a real value for f(x)) (The domain of f(x) is restricted to x>3))

3. Sep 26, 2004

### arildno

Note:
You've been asked to find the limit value when x approaches from one of the sides.
Do not confuse this with the more usual question whether the function has a limit at a point (that is, a unique number so that irrespective which side you approach from, your evaluations tend to that number)