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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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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. $

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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 $x=0$
answered Sep 24, 2013 by sreemathi.v
 

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