# Evaluate this limit and explain why it cannot be determined.

• Feb 22nd 2010, 04:19 PM
kmjt
Evaluate this limit and explain why it cannot be determined.
a) http://img21.imageshack.us/img21/595/alimit.th.png

b) Explain why the limit as x approaches 3 from the right cannot be determined.

c) What can you conclude about
http://img442.imageshack.us/img442/7...erlimit.th.png

For a I first tried to sub in 3 for x but that gives me the square root of 0 which is 0.. so is this indeterminite form?.. and i did it wrong?
• Feb 22nd 2010, 06:27 PM
drumist
For your purposes, it is probably most effective to just graph the function $f(x) = \sqrt{9-x^2}$ and notice what happens as you approach the point x=3 from the left and right sides.

By the way, $\sqrt{0} = 0$ is perfectly valid. It's not an indeterminate form. However, it *is* a borderline case. i.e., if you went any smaller than 0, then a problem arises. This is definitely related to the problem.
• Feb 22nd 2010, 06:34 PM
Prove It
Quote:

Originally Posted by kmjt
a) http://img21.imageshack.us/img21/595/alimit.th.png

b) Explain why the limit as x approaches 3 from the right cannot be determined.

c) What can you conclude about
http://img442.imageshack.us/img442/7...erlimit.th.png

For a I first tried to sub in 3 for x but that gives me the square root of 0 which is 0.. so is this indeterminite form?.. and i did it wrong?

For part a)

$\lim_{x \to 3^-}\sqrt{9 - x^2}$.

As said in a previous post, you should probably just graph the function and see what happens as $x \to 3$ from the left...

For part b)

Note that this function is only defined for $9 - x^2 \geq 0$

$9 \geq x^2$

$x^2 \leq 9$

$|x| \leq 3$

$-3 \leq x \leq 3$.

How can you make $x$ approach $3$ from the right if the function is not defined for any values of $x$ to the right of $3$?

c) Limits only exist when the left hand limit and right hand limit are equal.

Since there is a left hand limit but NOT a right hand limit, what does that tell you about the limit of this function?