Question

If  \[f\left( x \right) = \begin{cases}\frac{\cos^2 x - \sin^2 x - 1}{\sqrt{x^2 + 1} - 1}, & x \neq 0 \\ k , & x = 0\end{cases}\]   is continuous at x = 0, find k.

Answer 1

Given: 

\[f\left( x \right) = \binom{\frac{\cos^2 x - \sin^2 x - 1}{\sqrt{x^2 + 1} - 1}, x \neq 0}{k, x = 0}\] If f(x) is continuous at x = 0, then

\[\lim_{x \to 0} f\left( x \right) = f\left( 0 \right)\]
\[ \Rightarrow \lim_{x \to 0} \frac{\cos^2 x - \sin^2 x - 1}{\sqrt{x^2 + 1} - 1} = k\]
\[ \Rightarrow \lim_{x \to 0} \frac{1 - \sin^2 x - \sin^2 x - 1}{\sqrt{x^2 + 1} - 1} = k\]
\[ \Rightarrow \lim_{x \to 0} \frac{- 2 \sin^2 x}{\sqrt{x^2 + 1} - 1} = k\]
\[ \Rightarrow \lim_{x \to 0} \frac{- 2\left( \sin^2 x \right)\left( \sqrt{x^2 + 1} + 1 \right)}{\left( \sqrt{x^2 + 1} - 1 \right)\left( \sqrt{x^2 + 1} + 1 \right)} = k\]
\[ \Rightarrow \lim_{x \to 0} \frac{- 2\left( \ sin^2 x \right)\left( \sqrt{x^2 + 1} + 1 \right)}{x^2} = k\]
\[ \Rightarrow - 2 \lim_{x \to 0} \frac{\left( \ sin^2 x \right)\left( \sqrt{x^2 + 1} + 1 \right)}{x^2} = k\]
\[ \Rightarrow - 2 \lim_{x \to 0} \left( \frac{\ sinx}{x} \right)^2 \lim_{x \to 0} \left( \sqrt{x^2 + 1} + 1 \right) = k\]
\[ \Rightarrow - 2 \times 1 \times \left( 1 + 1 \right) = k\]
\[ \Rightarrow k = - 4\]