Question

# Find the value of $$k$$ for which $$f(x)=\begin{cases} \dfrac{1-\cos 4x}{8x^{2}}, when \ \ x\neq 0 \\ k\ \ ,\ \ \ \ \ \ \ \ \ \ \ \ \ when\ \ \ x=0 \end{cases}$$   is continuous at $$x=0$$.

Solution

## Given,$$f{\left( x \right)} = \begin{cases} \cfrac{1 - \cos{4x}}{8 {x}^{2}}, & \text{when } x \ne 0 \\ k, & \text{when } x = 0 \end{cases}$$$$f{\left( 0 \right)} = k$$If $$f{\left( x \right)}$$ is continuous at $$x = 0$$, then$$\displaystyle \lim_{x \rightarrow 0}{f{\left( x \right)}} = f{\left( 0 \right)}$$$$\displaystyle \lim_{x \rightarrow 0}{\cfrac{1 - \cos{4x}}{8 {x}^{2}}} = k$$$$\displaystyle \lim_{x \rightarrow 0}{\cfrac{2 \sin^{2}{2x}}{8 {x}^{2}}} = k$$$$\displaystyle \cfrac{2}{2} \lim_{x \rightarrow 0}{\cfrac{\sin^{2}{2x}}{4 {x}^{2}}} = k$$$$\displaystyle \Rightarrow \lim_{x \rightarrow 0}{\cfrac{\sin^{2}{2x}}{{\left( 2x \right)}^{2}}} = k$$$$\displaystyle \Rightarrow k = \lim_{x \rightarrow 0}{{\left( \cfrac{\sin{2x}}{2x} \right)}^{2}}$$$$\Rightarrow k = 1$$Mathematics

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