Question

If f (x) = x sin(1/x), x ~ 0, then lim f (x)=,
X->0
(a) 1
(b)0
(c)-1
(d) none​

Answer 1

$\large\underline{\sf{Solution-}}$

$\rm :\longmapsto\:\displaystyle \lim_{x \to \: 0} \: x \: sin\dfrac{1}{x}$

We know that

$\rm :\longmapsto\: - 1 \leqslant sinx \leqslant 1 \: \: where \: x \: is \: real$

Now,

$\rm :\longmapsto\:\displaystyle \lim_{x \to \: 0} \: x \: sin\dfrac{1}{x}$

$\rm \: \: = \: \: 0 \times (an \: oscillatory \: number \: lies \: between \: - 1 \: and \: 1)$

$\rm \: \: = \: \: 0$

$\: \: \: \: \: \: \: \: \: \: \: \: \boxed{\rm :\implies\:\displaystyle \lim_{x \to \: 0} \bf \: x \: sin\dfrac{1}{x} = 0}$

Hence,

$\boxed{\bf\implies \:Option \: (b) \: is \: correct}$

Additional Information :-

$\begin{gathered}(1)\:{\underline{\boxed{\bf{\blue{{\tt \:\lim_{x\to 0} \: \dfrac{sin \: x}{x} \: = \: 1 }}}}}} \\ \end{gathered}$

$\begin{gathered}(2)\:{\underline{\boxed{\bf{\blue{{\tt \:\lim_{x\to 0} \: \dfrac{tan \: x}{x} \: = \: 1 }}}}}} \\ \end{gathered}$

$\begin{gathered}(3)\:{\underline{\boxed{\bf{\blue{{\tt \:\lim_{x\to 0} \: \dfrac{ {e}^{x} \: - \: 1}{x} \: = \: 1 }}}}}} \\ \end{gathered}$

$\begin{gathered}(4)\:{\underline{\boxed{\bf{\blue{{\tt \:\lim_{x\to 0} \: \dfrac{ {a}^{x} \: - \: 1}{x} \: = \: log(a)}}}}}} \\ \end{gathered}$

$\begin{gathered}(5)\:{\underline{\boxed{\bf{\blue{{\tt \:\lim_{x\to 0} \: \dfrac{log(1 \: + \: x)}{x} \: = \: 1 }}}}}} \\ \end{gathered}$

$\begin{gathered}(6)\:{\underline{\boxed{\bf{\blue{{\tt \:\lim_{x\to 0} \: \dfrac{ {sin}^{ - 1}x }{x} \: = \: 1 }}}}}} \\ \end{gathered}$

$\begin{gathered}(7)\:{\underline{\boxed{\bf{\blue{{\tt \:\lim_{x\to 0} \: \dfrac{ {tan}^{ - 1}x }{x} \: = \: 1 }}}}}} \\ \end{gathered}$

$\begin{gathered}(8)\:{\underline{\boxed{\bf{\blue{{\tt \:\lim_{x\to 0} \: {(1 + x)}^{ \frac{1}{x} } \: = \: e }}}}}} \\ \end{gathered}$

$\begin{gathered}(9)\:{\underline{\boxed{\bf{\blue{{\tt \:\lim_{x\to \infty } \: {\bigg(1 + \dfrac{1}{x} \bigg) }^{x} \: = \: e }}}}}} \\ \end{gathered}$