Browse Questions

# Discuss the continuity of the following function at x = 0: $f(x) = \left\{ \begin{array}{l l}\large\frac{x^4+2x^3+x^2}{\tan^{-1}x}, & \quad if { x \; \neq 2 } \\ 0, & \quad if { x\; = 0 } \end{array} \right.$

Toolbox:
• If $f$ is a real function on a subset of the real numbers and $c$ be a point in the domain of $f$, then $f$ is continuous at $c$ if $\lim\limits_{\large x\to c} f(x) = f(c)$.
• $\lim\limits_{\large x\to 0}\large\frac{\tan^{-1}x}{x}$$=1 Step 1: f(x)=\left\{\begin{array}{1 1}\large\frac{x^4+2x^3+x^2}{\tan^{-1}x},&x\neq 2\\0,&x=0\end{array}\right. LHL at x\neq 2 \Rightarrow \lim\limits_{x\to 2^-}\large\frac{x^4+2x^3+x^2}{\tan^{-1}x} \Rightarrow \lim\limits_{x\to 2^-}\large\frac{x^2(x+1)^2}{\tan^{-1}x} \Rightarrow \lim\limits_{x\to 2^-}\large\frac{x(x+1)^2}{\Large\frac{\tan^{-1}x}{x}} Step 2: Since \lim\limits_{x\to 0}\large\frac{\tan^{-1}x}{x}$$=1$
$\Rightarrow \lim\limits_{x\to 2^-}\large\frac{2(2+1)^2}{1}$
$\Rightarrow 18$
Step 3:
Consider the RHL at $x=0$
$f(0)=\large\frac{0(0+1)^2}{1}$$=0$
Hence $f(0)=0$
Hence it is not continuous at $x=0$