求高數(shù)極限的題目及答案一、選擇題（每題5分，共20分）1.求極限\(\lim_{x\to0}\frac{\sinx}{x}\)的值是多少？A.0B.1C.2D.\(\infty\)答案：B2.求極限\(\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x\)的值是多少？A.0B.1C.\(e\)D.\(\infty\)答案：C3.求極限\(\lim_{x\to0}\frac{e^x-\cosx}{x^2}\)的值是多少？A.0B.1C.\(\frac{1}{2}\)D.\(\infty\)答案：C4.求極限\(\lim_{x\to1}\frac{x^2-1}{x-1}\)的值是多少？A.0B.1C.2D.\(\infty\)答案：C二、填空題（每題5分，共30分）1.求極限\(\lim_{x\to0}\frac{1-\cosx}{x^2}\)的值是\(\boxed{\frac{1}{2}}\)。2.求極限\(\lim_{x\to\infty}\frac{\lnx}{x}\)的值是\(\boxed{0}\)。3.求極限\(\lim_{x\to0}\frac{\tanx-\sinx}{x^3}\)的值是\(\boxed{\frac{1}{2}}\)。4.求極限\(\lim_{x\to0}\frac{e^x-1}{x}\)的值是\(\boxed{1}\)。5.求極限\(\lim_{x\to0}\frac{\ln(1+x)}{x}\)的值是\(\boxed{1}\)。6.求極限\(\lim_{x\to\infty}\frac{x^2+3x+2}{x^2-4x+4}\)的值是\(\boxed{1}\)。三、計(jì)算題（每題10分，共40分）1.求極限\(\lim_{x\to0}\frac{x^3-\sinx}{x^5}\)。解：利用洛必達(dá)法則，我們有\(zhòng)[\lim_{x\to0}\frac{x^3-\sinx}{x^5}=\lim_{x\to0}\frac{3x^2-\cosx}{5x^4}=\lim_{x\to0}\frac{6x+\sinx}{20x^3}=\lim_{x\to0}\frac{6+\cosx}{60x^2}=\lim_{x\to0}\frac{-\sinx}{120x}=\lim_{x\to0}\frac{-\cosx}{120}=\boxed{0}。\]2.求極限\(\lim_{x\to1}\frac{x^2-3x+2}{x^2-5x+6}\)。解：首先分解分子和分母，\[\lim_{x\to1}\frac{(x-1)(x-2)}{(x-2)(x-3)}=\lim_{x\to1}\frac{x-1}{x-3}=\frac{1-1}{1-3}=\boxed{-\frac{1}{2}}。\]3.求極限\(\lim_{x\to\infty}\left(\frac{x+1}{x-1}\right)^x\)。解：將表達(dá)式重寫為\[\lim_{x\to\infty}\left(1+\frac{2}{x-1}\right)^x=\lim_{x\to\infty}\left(1+\frac{2}{x-1}\right)^{x-1}\cdot\left(1+\frac{2}{x-1}\right)=e^2\cdot1=\boxed{e^2}。\]4.求極限\(\lim_{x\to0}\frac{1-\cos2x}{x^2}\)。解：利用倍角公式和極限的性質(zhì)，\[\lim_{x\to0}\frac{1-\cos2x}{x^2}=\lim_{x\to0}\frac{2\sin^2x}{x^2}=2\lim_{x\to0}\left(\frac{\sinx}{x}\right)^2=2\cdot1^2=\boxed{2}。\]四、證明題（每題10分，共10分）1.證明\(\lim_{x\to0}\frac{1-\cosx}{x^2}=\frac{1}{2}\)。證明：利用洛必達(dá)法則，\[\lim_{x\to0}\frac{1-\cosx}{x^2}=\lim