Browse Questions

# Examine the continuity of f(x) at the indicated point $f(x)=\left \{\begin{array}{1 1}\large\frac{2x^2-3x-2}{x-2} &\normalsize if\;x\neq 2\\5 & if\;x=2\end{array}\right.$ at x=2.

Toolbox:
• A function is said to be discontinuous at $x=a$ if both the RHL and the LHL does not exist.
• It is also discontinuous if RHL $\neq$ LHL.
Step 1:
$f(x)=\left \{\begin{array}{1 1}\large\frac{2x^2-3x-2}{x-2} &\normalsize if\;x\neq 2\\5 & if\;x=2\end{array}\right.$ at x=2.
Let $f(x)=\large\frac{2x^2-3x-2}{x-2}$
Step 2:
Consider $\large\frac{2x^2-3x-2}{x-2}=\frac{(2x+1)(x-2)}{x-2}$
$f(x)=2x+1$
Step 3:
The LHL at $x=2^-$
$\lim\limits_{\large x\to 2^-}f(x)=\lim\limits_{\large x\to 2^-}(2x+1)$
$\lim\limits_{\large h\to 0}2(2-h)+1=5$
Step 4:
Similarly the RHL at $x=2^+$
$\lim\limits_{\large x\to 2^+}f(x)=\lim\limits_{\large x\to 2^+}(2x+1)$
$\lim\limits_{\large h\to 0}2(2+h)+1=5$
Hence LHL=RHL.
Step5:
But at $x=2$
The function $f(x)=\large\frac{2x^2-3x-2}{x-2}$ is un defined.
Hence it does not exist.
Hence it is discontinuous.
edited Jun 25, 2013