Question
Let f(x) be a function defined by $\text{f(x)}=\begin{cases}\frac{3\text{x}}{|\text{x}|+2\text{x}}, & \text{x} \neq0\\\ \ \ \ 0, & \text{x} = 0\end{cases}.$

Answer

$\lim\limits_{\text{x}\rightarrow0^{-}}\text{f(x)}=\lim\limits_{\text{x}\rightarrow0^{-}}\frac{3\text{x}}{-\text{x}+2\text{x}}=\lim\limits_{\text{x}\rightarrow0^{-}}\frac{3\text{x}}{\text{x}}=3$ $[\because \text{x}\rightarrow^-,|\text{x}|=-\text{x}]$ $\lim\limits_{\text{x}\rightarrow0^+}\text{f(x)}=\lim\limits_{\text{x}\rightarrow0^+}\frac{3\text{x}}{\text{x}+2\text{x}}=1$ $[\because \text{x}\rightarrow0^+,|\text{x}|=\text{x}]$ Thus, $\lim\limits_{\text{x}\rightarrow0^{-}}\text{f(x)}\neq\lim\limits_{\text{x}\rightarrow0^{+}}\text{f(x)}$ $\therefore\ \lim\limits_{\text{x}\rightarrow0}\text{f(x)}$ does not exist.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free