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Prove that the function f defined by $ f(x)=\left \{\begin{array}{1 1}\large\frac{x}{\mid x\mid+2x^2}, & x\neq 0\\k, &x=0\end{array}\right.$remains discontinuous at x=0,regardless the choice of k.

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Toolbox:
  • A function is said to be discontinuous if the LHL $\neq$ RHL.
  • It is also said to be discontinuous if the function does not exist at both LHL and RHL.
Step 1:
Given : $ f(x)=\left \{\begin{array}{1 1}\large\frac{x}{\mid x\mid+2x^2}, & x\neq 0\\k, &x=0\end{array}\right.$
Now the LHL :
$\lim\limits_{\large x\to 0^-}f(x)=\lim\limits_{\large x\to 0^-}\large\frac{x}{\mid x\mid+2x^{\large 2}}$
Let $x=0-h$; as $x\to 0^-,h\to 0$
LHL =$\lim\limits_{\large h\to 0}\large\frac{-h}{\mid -h\mid+2(-h)^2}$
$\quad=\lim\limits_{\large h\to 0}\large\frac{-h}{h+2h^2}$
$\quad=\lim\limits_{\large h\to 0}\large\frac{-h}{h(1+2h)}$
On applying limits we get,
$\quad=\large\frac{-1}{1+2(0)}$
$\quad=-1$
Step 2:
The RHL :
$\lim\limits_{\large x\to 0^+}f(x)=\lim\limits_{\large x\to 0^+}\large\frac{x}{\mid x\mid+2x^{\large 2}}$
Put $x=0+h$; as $x\to 0^+,h\to 0$
RHL =$\lim\limits_{\large h\to 0}\large\frac{h}{\mid h\mid+2h^2}$
$\quad=\lim\limits_{\large h\to 0}\large\frac{h}{h+2h^2}$
$\quad=\lim\limits_{\large h\to 0}\large\frac{h}{h(1+2h)}$
On applying limits we get,
$\quad=\large\frac{1}{1+2(0)}$
$\quad=1$
LHL $\neq$ RHL
Hence $f(x)$ remains discontinuous at $x=0$,regardless of the choice of $k$
answered Jun 26, 2013 by sreemathi.v
 

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