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Given that $ f(x)= \left\{ \begin{array}{1 1} \Large\frac{1- \cos 4x}{x^2}, & \quad if\;x<0 \\ a & \quad ,if\;x=0 \\ \Large\frac{\sqrt x}{\sqrt{16+\sqrt{x}}-4}, & \quad if\;x>0 \end{array}. \right. $ If f(x) is continuous at x = 0, find the value of a.

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Toolbox:
  • $ LHL = RHL = f(0)$ for continuity
  • $ 1-cos4x=2sin^22x$
LHL
$ \lim\limits_{x \to 0} \large\frac{1-cos4x}{x^2}=8$
$ \lim\limits_{x \to 0} \large\frac{2sin^22x*4}{4x^2}=8$
RHL $ \lim\limits_{x \to 0} \large\frac{\sqrt x}{\sqrt{16}+\sqrt x-4}=8$
 
Rationalise denominator and proceed
$ \lim\limits_{x \to 0} \large\frac{\sqrt x [ \sqrt{16+\sqrt x}+4]}{\sqrt x}=8$
$ f $ is continuous $ \Rightarrow a = 8$

 

answered Mar 11, 2013 by thanvigandhi_1
edited Mar 26, 2013 by thanvigandhi_1
 

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