Question

If the function $f\left(x\right)$ is defined as:
$f\left(x\right)=\frac{x^4-64x}{\sqrt{x^2+9}-5}$, for $x\ne 4$
$=3$ ,for $x=4$
Show that $f\left(x\right)$ has a removable discontinuity at $x=4$.

Answer 1

$f\left(x\right)=\frac{x^4-64x}{\sqrt{x^2+9}-5}$, for $x\ne 4$

$=3$ ,for $x=4$

At $x\ne 4,f\left(x\right)=\frac{x(x^3-64)}{\sqrt{x^2+9}-5}=\frac{x(x^3-64)}{x^2+9-25}\left(\sqrt{x^2+9+5}\right)$

$=\frac{x(x-4)(x^2+16+4x)}{(x-4)(x+4)}\left(\sqrt{x^2+9+5}\right)$

$=\frac{x(x^2+16+4x)}{x+4}\left(\sqrt{x^2+9+5}\right)$

putting $x=4,f\left(x\right)=\frac{4(16+16+16)}{8}=\frac{3\left(16\right)\left(10\right)}{2}=240$

$f\left(x\right)$ can remove discontinuity by taking value 240 at x = 4

$f\left(x\right)=\frac{x^4-64x}{\sqrt{x^2+9}-5}$, for $x\ne 4$

$240$ ,for $x=4$

is continues thought out