question_answer

If\[f(x)=\left\{ \begin{matrix}    \frac{{{\cos }^{2}}x-{{\sin }^{2}}x-1}{\sqrt{{{x}^{2}}+1}-1}, & x\ne 0  \\    a, & x=0  \\ \end{matrix} \right.\] is continuous at x = 0, find the value of a.

Answer:

We have, \[\begin{matrix}    x\ne 0  \\    x=0  \\ \end{matrix}\] Since, f (x) is continuous at x = 0. \[\therefore \]                  \[\underset{x\to 0}{\mathop{\lim }}\,f\,(x)=f\,(0)\] \[\Rightarrow \]   \[\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{{{\cos }^{2}}x-{{\sin }^{2}}x-1}{\sqrt{{{x}^{2}}+1}-1} \right)=a\] \[\Rightarrow \]   \[\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{1-{{\sin }^{2}}x-{{\sin }^{2}}x-1}{\sqrt{{{x}^{2}}+1}-1} \right)=a\] \[\Rightarrow \]   \[\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{-\,2{{\sin }^{2}}x}{\sqrt{{{x}^{2}}+1-1}} \right)=a\] \[\Rightarrow \]\[-\,2\,\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{{{\sin }^{2}}x}{\sqrt{{{x}^{2}}+1-1}} \right)\times \frac{\sqrt{{{x}^{2}}+1}+1}{\sqrt{{{x}^{2}}+1}+1}=a\] \[\Rightarrow \]\[-\,2\,\,\underset{x\to 0}{\mathop{\lim }}\,{{\left( \frac{\sin x}{x} \right)}^{2}}\times \underset{x\to 0}{\mathop{\lim }}\,\sqrt{{{x}^{2}}+1}+1=a\] \[\left[ \because \underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x}{x}=1 \right]\] \[\Rightarrow \]               \[-\,2\,{{(1)}^{2}}\times (\sqrt{(10+1})+1)=a\] \[\Rightarrow \]                           \[-\,2\times 2=a\] \[\Rightarrow \]                           \[a=-\,4\]