Question

#Quality Question
@Continuity

Show that
$\displaystyle\sf f(x) =\begin{cases} \sf x \: sin \dfrac{1}{x} & \text{if } x \neq 0 \\ \\ 0 & \text{if }  \sf x = 0 \end{cases}\ \textless \ br /\ \textgreater$
is continuous at x = 0​

Answer 1

▪Given :-

$\bf f(x) =\begin{cases} \sf x \: sin \dfrac{1}{x} & \text{if } x \neq 0 \\ \\ 0 & \text{if }  \sf x = 0 \end{cases}$

To Prove :-

f(x) is continuous at x =  0

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▪Concept :-

$\sf If \:  \:  \:  f(x) =\begin{cases} \sf x \: sin\dfrac{1}{x} & \text{if } x \neq 0 \\ \\0 & \text{if }  \sf x = 0 \end{cases}$

is continuous at x = 0 then

$\sf\lim_{x \to 0}f(x) = f(0) = 0$

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▪Sandwich Theoram :-

If f , g and h are functions such that

$f(x) \leqslant g(x) \leqslant h(x)$

for all x in some neighborhood of c

and if

$\bf\lim_{x \to c}f(x) = p =  \bf\lim_{x \to c}h(x)$

Then,

$\bf\lim_{x \to c}g(x) = p$

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▪Proof :-

We know that

$\sf - 1 \leqslant sin \theta \leqslant 1$

So,

$\sf - 1 \leqslant sin \dfrac{1}{x}  \leqslant 1$

Mulyiplying by x

$\sf - x \leqslant x \: sin \dfrac{1}{x}   \leqslant x$

As

$\sf\lim_{x \to 0 {}^{ - } }( - x) = 0 =  \sf\lim_{x \to 0 {}^{ + } }x$

Hence

By Sandwich Theoram

$\bf\lim_{x \to 0}x \: sin \frac{1}{x}  = 0$

Hence

$\sf\lim_{x \to 0}f(x) = f(0) = 0$

So f(x) is continuous at x = 0 .

Answer 2

$\displaystyle\sf f(x) =\begin{cases} \sf x \: sin \dfrac{1}{x} & \text{if } x \neq 0 \\ \\ 0 & \text{if }  \sf x = 0$