# How do you evaluate sqrt(25-x^2) as x approaches 5-?

Apr 15, 2016

As $x \to 5 - , \sqrt{25 - {x}^{2}} \to 0$.

#### Explanation:

Unless otherwise stated, the graph of $y = \sqrt{25 - {x}^{2}}$ means $y = \pm \sqrt{25 - {x}^{2}}$ and represents the circle with center at the origin and radius = 5.

If it is indicated that $y = + \sqrt{25 - {x}^{2}}$, then the graph is the semi-circle above the x-axis, with dead-end-discontinuations, at $\left(\pm 5 , 0\right)$, and there exists this limit problem. . .

Apr 15, 2016

See below.

#### Explanation:

Here is the reasoning:

As $x$ approaches $5$, but $x$ is less than $5$,

${x}^{2}$ approaches $25$ and is less than $25$.

So $25 - {x}^{2}$ is positive and the square root is real.

Furthermore, $25 - {x}^{2}$ approaches $0$,

So, finally, $\sqrt{25 - {x}^{2}}$ approaches $0$

More formally,

${\lim}_{x \rightarrow {5}^{-}} \sqrt{25 - {x}^{2}} = \sqrt{{\lim}_{x \rightarrow {5}^{-}} \left(25 - {x}^{2}\right)}$

$= \sqrt{25 - {\lim}_{x \rightarrow {5}^{-}} \left({x}^{2}\right)}$

$= \sqrt{25 - {\left({\lim}_{x \rightarrow {5}^{-}} x\right)}^{2}}$

$= \sqrt{25 - {5}^{2}} = 0$