Question

How do you find the limit of `sqrt((4-x^2))` as x approaches `-2`?

Answer 1

It depends on how you have defined limits at the boundaries of domains..

Explanation:

The domain of `f(x) = sqrt(4-x^2)` is `[-2,2]`

There is no limit as `x` approaches `-2` from the left, because the function does not give real values for such `x`.
We do see that as `x` approaches `-2` from the right, we get values of `f(x)` approaching `0`.

If your definition has not provided for this kind of thing, we must say

`lim_(xrarr-2)sqrt(4-x^2)` `" "` does not exist.

For example, the definition in Stewart's Calculus is

Let `f` be defined on some open interval containing `c` except possible at `c` itself. Then `lim_(xrarrc)f(x) = L` if and only if

for every positive `epsilon`, there is a positive `delta` such that for all `x`, we have `0 < abs(x-c) < delta` implies `abs(f(x)-L) < epsilon`.

`f(x) = sqrt(4-x^2)` is not defined (real valued) on an open interval containing `-2`. So the limit as `x` approaches `-2` is not defined.

Note also, that in this definition, the conditional must hold for all `x` on either side of `c`.

Some treatments of the limit include in the definition a phrase that tells us to only consider values in the domain of `f` when evaluating limits.
They might say, for example
Let `f` be defined for some real numbers in an open interval containing `c`. Then `lim_(xrarrc)f(x) = L` if and only if;

for every positive `epsilon`, there is a positive `delta` such that for all `x` in the domain of `f`, we have `0 < abs(x-c) < delta` implies `abs(f(x)-L) < epsilon`.

For such a definition, we get

`lim_(xrarr-2)sqrt(4-x^2) = 0`