題目 Problem (10 points) Find the following limits. (a) limx→3x2+x−12x−3\lim_{x \to 3} \dfrac{x^2+x-12}{x-3}limx→3x−3x2+x−12. (b) limx→0x2secx−1\lim_{x \to 0} \dfrac{x^2}{\sec x-1}limx→0secx−1x2. 解答 解法一 (a) 分解 x2+x−12=(x+4)(x−3)x^2+x-12=(x+4)(x-3)x2+x−12=(x+4)(x−3)，故極限為 7\boxed{7}7。 (b) 以 secx−1=1−cosxcosx\sec x-1=\frac{1-\cos x}{\cos x}secx−1=cosx1−cosx，得 x2secx−1=cosx⋅x21−cosx⟶2.\frac{x^2}{\sec x-1} =\cos x\cdot\frac{x^2}{1-\cos x} \longrightarrow\boxed{2}.secx−1x2=cosx⋅1−cosxx2⟶2.
