Tính \(\lim\limits_{x\rightarrow0}\frac{\cos^4x-\sin^4x-1}{\sqrt{x^2+1}-1}\).
\(4\) \(-4\) \(2\) \(-2\) Hướng dẫn giải:Ta có \(\cos^4x-\sin^4x-1=\left(\cos^2x\right)^2-\left(\sin^2x\right)^2-1=\cos^2x-\sin^2x-1=-2\sin^2x\) và \(\sqrt{x^2+1}-1=\frac{\left(x^2+1\right)-1}{\sqrt{x^2+1}+1}\) nên \(\lim\limits_{x\rightarrow0}\frac{\cos^4x-\sin^4x-1}{\sqrt{x^2+1}-1}\)\(=\lim\limits_{x\rightarrow0}\frac{-2\sin^2x.\left(\sqrt{x^2+1}+1\right)}{x^2}=\lim\limits_{x\rightarrow0}\left(-2\left(\frac{\sin x}{x}\right)^2\left(\sqrt{x^2+1}+1\right)\right)=-4.\)