Question

ans this question solve it​

Answer 1

$\\ \large{ \underline{ \underline{ \bigstar \: \: \: \: \: { \pmb{ \sf{Solution \: : }}}}}}$

$\sf \: f(x) = \begin{cases} \sf \frac{2x - |x| }{x}, & \sf \: \: \: \: x \not = 0 \\ \\ \\ \: \: \: \: \sf 2 \: \: \: \: ,& \sf\: \: \: \: x = 0 \end{cases}$

Function exist at x = 0 if

$\dashrightarrow \sf f( {0}^{ + } ) = f( {0}^{ - } ) = f(0)$

$\begin{array}{c|c| c} \dashrightarrow \sf \lim_{x \to {0}^{ + } } f(x)& \sf \dashrightarrow \lim_{x \to {0}^{ - } } f(x)& \dashrightarrow \sf \: f(x)| _{x=0} = 2 \\ \\ \dashrightarrow \sf\lim_{x \to {0}^{ + }} \bigg( \frac{2x - |x| }{x} \bigg) & \dashrightarrow\sf\lim_{x \to {0}^{ - }} \bigg( \frac{2x - |x| }{x} \bigg)&\sf \dashrightarrow f(x)| _{x=0} = 2 \\ \\ \dashrightarrow \sf\lim_{x \to {0}^{ + } } \bigg(2 - \frac{ |x| }{x} \bigg)& \dashrightarrow \sf\lim_{x \to {0}^{ - } } \bigg(2 - \frac{ |x| }{x} \bigg)& \dashrightarrow\sf \: f(x)| _{x=0} = 2 \\ \\ \dashrightarrow\sf\lim_{x \to {0}^{ + } } \bigg(2 - \frac{ {x}^{ } }{ {x}^{ } } \bigg)& \dashrightarrow\sf\lim_{x \to {0}^{ - } } \bigg(2 + \frac{x}{x} \bigg) & \sf \dashrightarrow \: f(x)| _{x=0} = 2 \\ \\ \dashrightarrow\sf\lim_{x \to {0}^{ + } } \bigg(2 - 1 \bigg)& \dashrightarrow\sf\lim_{x \to {0}^{ - } } \bigg(2 + 1 \bigg)& \sf \dashrightarrow \: f(x)| _{x=0} = 2 \\ \\ \dashrightarrow \sf \: 1& \dashrightarrow \sf 3& \dashrightarrow \sf2\end{array} \\ \\ \\ \bf \therefore \: \: \: Limit \: \: doesn't \: \: exist \: \: at \: \: x= 0$