Question

If $f\left(x\right)=\left\{\begin{array}{ll}\frac{{\cos }^{2}x-{\sin }^{2}x-1}{\sqrt{x^2+1}-1},& x\ne 0\\ k,& x=0\end{array}\right.$ is continuous at x = 0, find k.

Answer 1

Given:
$f\left(x\right)=\left\{\begin{array}{l}\frac{{\cos }^{2}x-{\sin }^{2}x-1}{\sqrt{x^2+1}-1},x\ne 0\\ k,x=0\end{array}\right.$

If f(x) is continuous at x = 0, then

​$$\begin{gathered} \underset{x\to 0}{\lim }f\left(x\right)=f\left(0\right) \\ \Rightarrow \underset{x\to 0}{\lim }\frac{{\cos }^{2}x-{\sin }^{2}x-1}{\sqrt{x^2+1}-1}=k \\ \Rightarrow \underset{x\to 0}{\lim }\frac{1-{\sin }^{2}x-{\sin }^{2}x-1}{\sqrt{x^2+1}-1}=k \\ \Rightarrow \underset{x\to 0}{\lim }\frac{-2{\sin }^{2}x}{\sqrt{x^2+1}-1}=k \\ \Rightarrow \underset{x\to 0}{\lim }\frac{-2\left({\sin }^{2}x\right)\left(\sqrt{x^2+1}+1\right)}{\left(\sqrt{x^2+1}-1\right)\left(\sqrt{x^2+1}+1\right)}=k \\ \Rightarrow \underset{x\to 0}{\lim }\frac{-2\left({\sin }^{2}x\right)\left(\sqrt{x^2+1}+1\right)}{x^2}=k \\ \Rightarrow -2\underset{x\to 0}{\lim }\frac{\left({\sin }^{2}x\right)\left(\sqrt{x^2+1}+1\right)}{x^2}=k \\ \Rightarrow -2\underset{x\to 0}{\lim }{\left(\frac{\sin x}{x}\right)}^2\underset{x\to 0}{\lim }\left(\sqrt{x^2+1}+1\right)=k \\ \Rightarrow -2\times 1\times \left(1+1\right)=k \\ \Rightarrow k=-4 \end{gathered}$$