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Find whether $f(x)$ is continuous or discontinuous at the indicated point $f(x)=\left \{\begin{array}{1 1}( x-a)\sin\large\frac{1}{x-a},\normalsize & if\;x\neq 0\\0, & if\;x=0\end{array}\right.$ at $x=a$.

$\begin{array}{1 1} \text{Continuous} \\ \text{Discontinuous }\end{array} $

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Toolbox:
  • A function is said to continuous at a point if the LHL=RHL.
  • (i.e) $\lim\limits_{\large x\to a^+}f(x)=\lim\limits_{\large x\to a^-}f(x)$
Step 1:
Given : $f(x)=\left \{\begin{array}{1 1}( x-a)\sin\large\frac{1}{x-a},\normalsize & if\;x\neq 0\\0, & if\;x=0\end{array}\right.$ at x=a.
We have $f(x)=(x-a)\sin\large\frac{1}{x-a}$ when $x\neq a$
Now let us find the left hand limit.
$\lim\limits_{\large x\to a^-}f(x)=\lim\limits_{\large h\to a}f(a-h)$
$\qquad\qquad=\lim\limits_{\large h\to a}(a-h-a)\sin\large\frac{1}{(a-h-a)}$
$\qquad\qquad=\lim\limits_{\large h\to a}(-h)\sin\large\frac{1}{(-h)}$
But $\sin(-\theta)=-\sin\theta$
$\qquad\qquad=\lim\limits_{\large h\to a}(-h)(-\sin\large\frac{1}{(h)})$
$\qquad\qquad=\lim\limits_{\large h\to a}h\big(\sin\large\frac{1}{h}\big)$
Step 2:
As $h\to a,\large\frac{1}{h}$$\to \large\frac{1}{a}$
On applying limits we get
$\qquad\qquad=\lim\limits_{\large h\to a}a\sin\big(\large\frac{1}{a}\big)$
$\qquad\qquad=\lim\limits_{\large h\to a}\bigg(\large\frac{\sin\Large\frac{1}{a}}{\Large\frac{1}{a}}\bigg)$
But $\large\frac{\sin\theta}{\theta}$$=1$
$\qquad\qquad=1$
Step 3:
Next let us find the right hand limit.
$\lim\limits_{\large x\to a^+}f(x)=\lim\limits_{\large h\to a}f(a+h)$
$\qquad\qquad=\lim\limits_{\large h\to a}(a+h-a)\sin\large\frac{1}{(a+h-a)}$
$\qquad\qquad=\lim\limits_{\large h\to a}h\big(\sin\large\frac{1}{h}\big)$
Step 4:
As $h\to a,\large\frac{1}{h}$$\to \large\frac{1}{a}$
On applying limits we get
$\qquad\qquad=\lim\limits_{\large h\to a}a\sin\big(\large\frac{1}{a}\big)$
$\qquad\qquad=\lim\limits_{\large h\to a}\bigg(\large\frac{\sin\Large\frac{1}{a}}{\Large\frac{1}{a}}\bigg)$
$\qquad\qquad=1$
Hence LHL = RHL.
Step 5:
$f(a)=(a-a)\sin\large\frac{1}{a-a}$
$\qquad=0$
Hence it is continuous at $x=a$
answered Jun 26, 2013 by sreemathi.v
 

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