The properties of function f(x) as 1) The domain of f is (-\infty,\inft

Step 1
Explanation:
Consider the function
$f(x)=\{\begin{array}{ll}\frac{x(x-3)+70}{x^2+1}& ifx\ne 4\\ 15& ifx=4\end{array}$
The above function is defined over all real number therefore the domain of the function is ${\displaystyle (-\mathrm{\infty },\mathrm{\infty })}$,
Hence the property (1) is satisfied.
Evaluating ${\displaystyle f\left(3\right)}$,
${\displaystyle f\left(3\right)=\frac{3(3-3)+70}{3^2+1}=\frac{70}{10}=7}$
Hence the property (2) is satisfied.
Step 2
Evaluating ${\displaystyle \underset{x\to 2^{+}}{lim}f\left(x\right),}$
${\displaystyle \underset{x\to 2^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(2+h)}$
${\displaystyle Rightarrwo\underset{x\to 2^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}\frac{(2+h)(2+h-3)+70}{{(2+h)}^2+1}}$
${\displaystyle \Rightarrow \underset{x\to 2^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}\frac{(2+h)(h-1)+70}{{(2+h)}^2+1}}$
${\displaystyle \Rightarrow \underset{x\to 2^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}\frac{h^2+h+68}{h^2+4h+5}}$
${\displaystyle \Rightarrow \underset{x\to 2^{+}}{lim}f\left(x\right)=\frac{68}{5}}$
Evaluating ${\displaystyle \underset{x\to 2^{-}}{lim}f\left(x\right),}$
${\displaystyle \underset{x\to 2^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(2-h)}$
${\displaystyle \Rightarrow \underset{x-2^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}\frac{(2-h)(2-h-3)+70}{{(2+h)}^2+1}}$
${\displaystyle \Rightarrow \underset{x-2^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}\frac{(2-h)(-h-1}{}}$