[연습문제] 극한
1. 다음의 극한값 을 구하여라. (1) \(\displaystyle\lim_{x\to2}\frac{x^3+2x^2-3x-4}{x+2}=\frac{2^3+(2)2^2-(3)(2)-4}{2+2}={3\over2}\) (2) \(\displaystyle\lim_{x\to-2}\frac{x+2}{x^2-4x-12}=\lim_{x\to-2}{1\over x-6}=-{1\over8}\) (3) \(\displaystyle\lim_{x\to0}\frac{2x^2-x}{x^2-3x}=\lim_{x\to0}\frac{2x-1}{x-3}={1\over3}\) (4) \(\displaystyle\lim_{x\to0}\frac{x}{\tan2x}=\lim_{x\to0}\frac{2x}{2\tan2x}={1\over2}\) (5) \(\displaystyle\lim_{x\to0}\frac{1-\cos x}{\sin^2x}=\lim_{x\to0}\frac{1}{1+\cos x}={1\over2}\) (6) \(\displaystyle\lim_{x\to\infty}\frac{4x^3-5x^2+1}{x-x^2-7x^3}=\lim_{x\to\infty}\frac{4-{5\over x}+{1\over x^2}}{{1\over x^2}-{1\over x}-7}=-{4\over7}\) (7) \(\displaystyle\lim_{x\to1}\frac{x^n-1}{x-1}\text{(n은 자연수)}=\lim_{x\to1}(1+x+x^2+\cdots+x^{n-1})=n\) 위의 문제는 등비수열 을 참조한다. (8) \(\displaystyle\lim_{x\to\infty}\left(\sqrt{x+1}-\sqrt{x}\right)=\lim_{x\to\infty}\frac{1}{\sqrt{x+1}+\sqrt{x}}=0\) (9) \(\displaystyle\lim_{x\to\infty}x\left(x-\sqrt{x^2-a^2}\right)=\lim_{x\to\infty}\frac{a^2x}{