问题及解答

求极限 $\lim\limits_{x\rightarrow 0}\dfrac{\sec x-\cos x}{x^2}$.

\[
\lim_{x\rightarrow 0}\dfrac{\sec x-\cos x}{x^2}=\lim_{x\rightarrow 0}\dfrac{\frac{1}{\cos x}(1-\cos^2 x)}{x^2}=\lim_{x\rightarrow 0}\dfrac{\sin^2 x}{x^2}=\lim_{x\rightarrow 0}\dfrac{x^2}{x^2}=1.
\]