# Solve the following systems of equations:

Question:

$f(x)=\left\{\begin{array}{ll}\frac{2 x^{2}-3 x-2}{x-2} & \text {, if } x \neq 2 \\ 5, & \text { if } x=2\end{array}\right.$ at $\mathrm{x}=2$

Solution:

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.