Let $f(x)=\begin{cases} \frac{x^4-5x^2+4}{|(x-1)(x-2)|}, & x\neq1,2 \\ 6, & x=1 \\ 12, & x=2 \end{cases}$. Then $f(x)$ is continuous on the set:
R - {1, 2}
R
R - {1}
R - {2}
Solution
Solution:
Given:
$f(x)=\begin{cases}\frac{x^4-5x^2+4}{|(x-1)(x-2)|},&x\neq1,2\\6, &x=1\\12,&x=2\end{cases}$
Now,
$\begin{aligned}x^4-5x^2+4&=x^4-x^2-4x^2+4\\&=x^2(x^2-1)-4(x^2-1)\end{aligned}$
$=(x^2-1)(x^2-4)=(x-1)(x+1)(x-2)(x+2)$
$\Rightarrow f(x)=\begin{cases}\frac{(x-1)(x+1)(x-2)(x+2)}{|(x-2)(x-1)|},&x\neq1,2\\6,&x=1\\12,&x=2\end{cases}$
$\Rightarrow f(x)=\begin{cases}(x+1)(x+2),&x<1\\-(x+1)(x+2),&12\\6,&x=1\\12,&x=2\end{cases}$
So,
$\begin{aligned}\lim\limits_{x\rightarrow1^-}f(x)&=\lim\limits_{h\rightarrow0}f(1-h)\\&=\lim\limits_{h\rightarrow0}(1-h+1)(1-h+2)\\&=2\times3=6\end{aligned}$
$\begin{aligned}\lim\limits_{x\rightarrow1^+}f(x)&=-\lim\limits_{h\rightarrow0}(1+h+1)(1+h+2)\\&=-2\times3=-6\end{aligned}$
Also,
$\begin{aligned}\lim\limits_{x\rightarrow2^-}f(x)&=-\lim\limits_{h\rightarrow0}(2-h+1)(2-h+2)\\&=-12\end{aligned}$
$\begin{aligned}\lim\limits_{x\rightarrow2^+}f(x)&=\lim\limits_{h\rightarrow0}(2+h+1)(2+h+2)\\&=12\end{aligned}$
Thus,
$\lim\limits_{x\rightarrow1^+}f(x)\neq\lim\limits_{x\rightarrow1^-}f(x)$ and $\lim\limits_{x\rightarrow2^+}f(x)\neq\lim\limits_{x\rightarrow2^-}f(x)$
Therefore, the only points of discontinuity of the function $f(x)$ are $x = 1$ and $x = 2$. Hence, the given function is continuous on the set $R - \{1, 2\}$.