Question

# Solve: $$\lim _ { x \rightarrow 0 } \frac { 1 - \cos x } { x }$$

Solution

## $${\lim}_{x\rightarrow\,0}\dfrac{1-\cos{x}}{x}$$$$LHL={\lim}_{h\rightarrow\,{0}^{-}}\dfrac{1-\cos{\left(-h\right)}}{-h}$$$$={\lim}_{h\rightarrow\,{0}^{-}}\dfrac{1-\cos{h}}{-h}$$$$={\lim}_{h\rightarrow\,{0}^{-}}\dfrac{1-1+2{\sin}^{2}{\dfrac{h}{2}}}{-h}$$$$={\lim}_{h\rightarrow\,{0}^{-}}\dfrac{2{\sin}^{2}{\dfrac{h}{2}}}{-{\left(\dfrac{h}{2}\right)}^{2}h}\times {\left(\dfrac{h}{2}\right)}^{2}$$$$={\lim}_{h\rightarrow\,{0}^{-}}\dfrac{-2}{4}\times h=-\dfrac{1}{2}\times 0=0$$$$RHL={\lim}_{h\rightarrow\,{0}^{+}}\dfrac{1-\cos{\left(h\right)}}{h}$$$$={\lim}_{h\rightarrow\,{0}^{+}}\dfrac{1-\cos{h}}{-h}$$$$={\lim}_{h\rightarrow\,{0}^{+}}\dfrac{1-1+2{\sin}^{2}{\dfrac{h}{2}}}{h}$$$$={\lim}_{h\rightarrow\,{0}^{+}}\dfrac{2{\sin}^{2}{\dfrac{h}{2}}}{{\left(\dfrac{h}{2}\right)}^{2}h}\times {\left(\dfrac{h}{2}\right)}^{2}$$$$={\lim}_{h\rightarrow\,{0}^{+}}\dfrac{2}{4}\times h=\dfrac{1}{2}\times 0=0$$$$LHL=RHL=0$$Hence $${\lim}_{x\rightarrow\,0}\dfrac{1-\cos{x}}{x}=0$$Mathematics

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