Question

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

Answer 1

As `xto5-, sqrt(25-x^2)to0`.

Explanation:

Unless otherwise stated, the graph of `y = sqrt(25-x^2)` means `y=+-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 `(+-5, 0)`, and there exists this limit problem. . .

Answer 2

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_(xrarr5^-)sqrt(25-x^2) = sqrt(lim_(xrarr5^-)(25-x^2))`

`= sqrt(25-lim_(xrarr5^-)(x^2))`

`= sqrt(25-(lim_(xrarr5^-)x)^2)`

`= sqrt(25-5^2) = 0`