Question

Explain why the function is discontinuous at the given number

Answer 1

For $f$ to be continuous at $x=5$, we need to have

$\displaystyle\lim_{x\to5}f(x)=f(5)$

Note that $x\to5$ means that $x\neq5$, but that $x$ is *approaching* 5. We're told that for $x\neq5$, we have

$f(x)=\dfrac{x^2-5x}{x^2-25}$

We can write

$\dfrac{x^2-5x}{x^2-25}=\dfrac{x(x-5)}{(x+5)(x-5)}=\dfrac x{x+5}$

and the limit would be

$\displaystyle\lim_{x\to5}\frac x{x+5}=\dfrac5{5+5}=\dfrac5{10}=\dfrac12\neq1=f(5)$

and so $f$ is discontinuous.