Something as simple as this

$For\quad small\quad angle\quad Sin(x)\approx x\\ \lim _{ x\rightarrow { 0 } }{ (\frac { 1 }{ x } +\frac { 1 }{ x } } )=\lim _{ x\rightarrow { 0 } }{ (\frac { 2 }{ x } } )\\ R.H.L.=\lim _{ x\rightarrow { 0 }^{ + } }{ (\frac { 1 }{ x } +\frac { 1 }{ \sin { x } } } )=(\frac { 1 }{ { 0 }^{ + } } +\frac { 1 }{ { 0 }^{ + } } )=+\infty \\ L.H.L.=\lim _{ x\rightarrow { 0 }^{ - } }{ (\frac { 1 }{ { 0 }^{ - } } +\frac { 1 }{ { 0 }^{ - } } } )=-\infty \\ R.H.L.\neq L.H.L.\\ =>\quad Limit\quad does\quad not\quad exist\\$ Graph

$\lim_{x \to 0} (\frac{1}{x} + \frac{1}{sinx}) =\lim_{x \to 0} (\frac{x + sin x}{x \cdot sinx}) =$ $\lim_{x \to 0} \frac{2x}{x^2} = \lim_{x \to 0} \frac{2}{x}$ This limit doesn't exist because $\lim_{x \to 0^{+}} \frac{2}{x} = +\infty \text{ and } \lim_{x \to 0^{-}} \frac{2}{x} = -\infty$