Question

Discuss the continuity of the following function at $x=1$
$F\left(x\right)=\{\begin{array}{c}\frac{1}{1-x}-\frac{3}{1-x^2}+\frac{7}{4};\text{when}x<1\\ \frac{3}{4};\text{when}x=1\\ \frac{\log x}{x-1}-\frac{1}{4};\text{when}x>1\end{array}$

Answer 1

$f\left(x\right)=\frac{1}{1-x}-\frac{3}{1-x^3}+\frac{7}{4}$ when $x<1$

$=\frac{1}{1-x}[1-\frac{3}{1+x+x^2}]+\frac{7}{4}$

$=\frac{1}{1-x}\left[\frac{1+x+x^2-3}{1+x+x^2}\right]+\frac{7}{4}$

$=\frac{1}{1-x}\left[\frac{x^2+x-2}{1+x+x^2}\right]+\frac{7}{4}$

$=\frac{1}{1-x}\left[\frac{(x-1)(x+2)}{x^2+x+1}\right]+\frac{7}{4}$

$=-\left[\frac{(x+2)}{x^2+x+1}\right]+\frac{7}{4}$

$\therefore LHL=\underset{x\to 1^{-1}}{lim}-\left[\frac{(x+2)}{x^2+x+1}\right]+\frac{7}{4}$

$=-\left[\frac{1+2}{1+1+1}\right]=\frac{7}{4}$

$=-\frac{3}{3}+\frac{7}{4}=\frac{3}{4}$
$g\left(x\right)=\frac{\log x}{x-1}-\frac{1}{4}=\frac{4\log x-x+1}{4x-4}$
$\therefore RHL=\underset{x\to 1^{+1}}{lim}\frac{4\log x-x+1}{4x-4}$
$=\underset{x\to 1^{+1}}{lim}\frac{\frac{4}{x}-1}{4}\text{[using L'hospital's rule]}=\frac{\left(4/1\right)-1}{4}=\frac{3}{4}$
and $F\left(1\right)=\frac{3}{4}$

$LHL=F\left(1\right)=RHL$ at $x=1$
$\therefore F\left(x\right)$ is continuous at $x=1$