question_answer

For what value of k, is the function Continuous at   x = 0?

Answer:

Given function is Also, given f (x) is continuous at x = 0. \[\therefore \]      \[{{(LHL)}_{x\,=\,0}}={{(RHL)}_{x\,=\,0}}=f\,(0)\] Now,     \[LHL=\underset{x\,\to \,0}{\mathop{\lim }}\,f\,(x)\] \[=\underset{x\,\to \,{{0}^{-}}}{\mathop{\lim }}\,\frac{1-\cos 4x}{8{{x}^{2}}}=\underset{h\,\to \,0}{\mathop{\lim }}\,\frac{1-\cos \,(-\,Ah)}{8{{h}^{2}}}\] \[[put\,\,x=0-h=-\,h,\,\,when\,\,x\to 0,\,\,h\to 0]\] \[=\underset{h\,\to \,{{0}^{{}}}}{\mathop{\lim }}\,\frac{1-\cos 4h}{8{{x}^{2}}}=\underset{h\,\to \,0}{\mathop{\lim }}\,\frac{2{{\sin }^{2}}2h}{8{{h}^{2}}}\] \[\Rightarrow \]   \[\underset{h\,\to \,0}{\mathop{\lim }}\,\frac{{{\sin }^{2}}2h}{4{{h}^{2}}}=\underset{h\,\to \,0}{\mathop{\lim }}\,{{\left( \frac{\sin 2h}{2h} \right)}^{2}}=1\] At \[x=0,\]\[f\,(0)=k\] Now, from Eq. (i), we get \[LHL=f\,(0)\Rightarrow 1.1=k\Rightarrow k=1\] Hence, for \[k=1\] the given function \[f\,(x)\]is continuous at\[x=0\].