Question

Discuss the continuity of the following function at $x=1$.
$f\left(x\right)=\{\begin{array}{c}\frac{1}{1-x}-\frac{3}{1-x^3}+\frac{7}{4}\text{when}x<1\\ \\ \frac{3}{4}\text{when}x=1\\ \\ \frac{\log x}{x-1}-\frac{1}{4}\text{when}x>1\end{array}$

Answer 1

At $x=1$

L.H.L $=\underset{x\to 1-h,x\to 1}{lim}(\frac{1}{1-x}-\frac{3}{1-x^3}+\frac{7}{4})$

$=\underset{h\to 0}{lim}(\frac{1}{1-(1-h)}-\frac{3}{1-(1-h{)}^3}+\frac{7}{4})$

$=\underset{h\to 0}{lim}(\frac{1}{h}-\frac{3}{h^3+3h-3h^2}+\frac{7}{4})$

$=\underset{h\to 0}{lim}\left\{\frac{1}{h}(\frac{1}{1}-\frac{3}{h^2+3-3h}+\frac{7}{4})\right\}$

$=\underset{h\to 0}{lim}\{\frac{1}{h}\left(\frac{h(h-3)}{h^2+3-3h}\right)+\frac{7}{4}\}$

$=\frac{0-3}{0+3-0}+\frac{7}{4}$

$=-1+\frac{7}{4}=\frac{3}{4}$

R.H.L. $=\underset{x\to 1+h,x\to 1}{lim}\frac{\log x}{x-1}-\frac{1}{4}$

$=\underset{h\to 0}{lim}\frac{\log (1+h)}{1+h-1}-\frac{1}{4}$

$=\underset{h\to 0}{lim}\frac{\log (1+h)}{h}-\frac{1}{4}$

$=1-\frac{1}{4}\because \underset{x\to 0}{lim}\frac{\log (1+x)}{x}=1$

$=\frac{3}{4}$

L.H.L. $=$ R.H.L. $=f\left(0\right)$

So, $f\left(x\right)$ is continuous at $x=1$.