$f(x)=\left\{\begin{array}{ll}\frac{2 x^{2}-3 x-2}{x-2} & \text {, if } x \neq 2 \\ 5, & \text { if } x=2\end{array}\right.$ at $\mathrm{x}=2$
The given fucntion at $x \neq 0$ can be rewritten as,
$f(x)=\frac{2 x^{2}-3 x-2}{x-2}$
$=\frac{2 x^{2}-4 x+x-2}{x-2}=\frac{2 x(x-2)+1(x-2)}{x-2}$
$=\frac{(2 x+1)(x-2)}{x-2}=2 x+1$
Now,
$\lim _{x \rightarrow 2^{-}} f(x)=2 x+1$
$=\lim _{h \rightarrow 0} 2(2-h)+1=4+1=5$
$\lim _{x \rightarrow 2^{+}} f(x)=2 x+1$
$=\lim _{h \rightarrow 0} 2(2+h)+1=4+1=5$
$\lim _{x \rightarrow 2} f(x)=5$
As $\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2} f(x)=5$
The given fucntion at $x \neq 0$ can be rewritten as,
$f(x)=\frac{2 x^{2}-3 x-2}{x-2}$
$=\frac{2 x^{2}-4 x+x-2}{x-2}=\frac{2 x(x-2)+1(x-2)}{x-2}$
$=\frac{(2 x+1)(x-2)}{x-2}=2 x+1$
Now,
$\lim _{x \rightarrow 2^{-}} f(x)=2 x+1$
$=\lim _{h \rightarrow 0} 2(2-h)+1=4+1=5$
$\lim _{x \rightarrow 2^{+}} f(x)=2 x+1$
$=\lim _{h \rightarrow 0} 2(2+h)+1=4+1=5$
$\lim _{x \rightarrow 2} f(x)=5$
As $\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2} f(x)=5$
Thus, f(x) is continuous at x = 2.
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.