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

Mar 30, 2016

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

#### Explanation:

The domain of $f \left(x\right) = \sqrt{4 - {x}^{2}}$ is $\left[- 2 , 2\right]$

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 $f \left(x\right)$ approaching $0$.

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

${\lim}_{x \rightarrow - 2} \sqrt{4 - {x}^{2}}$ $\text{ }$ 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 ${\lim}_{x \rightarrow c} f \left(x\right) = L$ if and only if

for every positive $\epsilon$, there is a positive $\delta$ such that for all $x$, we have $0 < \left\mid x - c \right\mid < \delta$ implies $\left\mid f \left(x\right) - L \right\mid < \epsilon$.

$f \left(x\right) = \sqrt{4 - {x}^{2}}$ 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 ${\lim}_{x \rightarrow c} f \left(x\right) = L$ 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 $0 < \left\mid x - c \right\mid < \delta$ implies $\left\mid f \left(x\right) - L \right\mid < \epsilon$.

For such a definition, we get

${\lim}_{x \rightarrow - 2} \sqrt{4 - {x}^{2}} = 0$