题目
1.按定义证明下列极限:-|||-(1) lim _(xarrow +infty )dfrac (6x+5)(x)=6;-|||-(2) lim _(xarrow 2)((x)^2-6x+10)=2;-|||-(3) lim _(xarrow infty )dfrac ({x)^2-5}({x)^2-1}=1;-|||-(4) lim _(xarrow {2)^-}sqrt (4-{x)^2}=0;-|||-(5) lim cos x=cos (x)_(0)-|||-x→x0
题目解答
答案
解析
(1) $\lim _{x\rightarrow +\infty }\dfrac {6x+5}{x}=6$
步骤 1：将分子分母同时除以 $x$
$$\lim _{x\rightarrow +\infty }\dfrac {6x+5}{x} = \lim _{x\rightarrow +\infty }\dfrac {6+\frac{5}{x}}{1}$$
步骤 2：当 $x\rightarrow +\infty$ 时，$\frac{5}{x}\rightarrow 0$
$$\lim _{x\rightarrow +\infty }\dfrac {6+\frac{5}{x}}{1} = 6$$
(2) $\lim _{x\rightarrow 2}({x}^{2}-6x+10)=2$
步骤 1：直接代入 $x=2$
$$\lim _{x\rightarrow 2}({x}^{2}-6x+10) = 2^2 - 6\cdot2 + 10 = 4 - 12 + 10 = 2$$
(3) $\lim _{x\rightarrow \infty }\dfrac {{x}^{2}-5}{{x}^{2}-1}=1$
步骤 1：将分子分母同时除以 ${x}^{2}$
$$\lim _{x\rightarrow \infty }\dfrac {{x}^{2}-5}{{x}^{2}-1} = \lim _{x\rightarrow \infty }\dfrac {1-\frac{5}{{x}^{2}}}{1-\frac{1}{{x}^{2}}}$$
步骤 2：当 $x\rightarrow \infty$ 时，$\frac{5}{{x}^{2}}\rightarrow 0$ 和 $\frac{1}{{x}^{2}}\rightarrow 0$
$$\lim _{x\rightarrow \infty }\dfrac {1-\frac{5}{{x}^{2}}}{1-\frac{1}{{x}^{2}}} = 1$$
(4) $\lim _{x\rightarrow {2}^{-}}\sqrt {4-{x}^{2}}=0$
步骤 1：当 $x\rightarrow {2}^{-}$ 时，$4-{x}^{2}\rightarrow 0$
$$\lim _{x\rightarrow {2}^{-}}\sqrt {4-{x}^{2}} = \sqrt{0} = 0$$
(5) $\lim _{x\rightarrow 0}\cos x=\cos 0$
步骤 1：直接代入 $x=0$
$$\lim _{x\rightarrow 0}\cos x = \cos 0 = 1$$
步骤 1：将分子分母同时除以 $x$
$$\lim _{x\rightarrow +\infty }\dfrac {6x+5}{x} = \lim _{x\rightarrow +\infty }\dfrac {6+\frac{5}{x}}{1}$$
步骤 2：当 $x\rightarrow +\infty$ 时，$\frac{5}{x}\rightarrow 0$
$$\lim _{x\rightarrow +\infty }\dfrac {6+\frac{5}{x}}{1} = 6$$
(2) $\lim _{x\rightarrow 2}({x}^{2}-6x+10)=2$
步骤 1：直接代入 $x=2$
$$\lim _{x\rightarrow 2}({x}^{2}-6x+10) = 2^2 - 6\cdot2 + 10 = 4 - 12 + 10 = 2$$
(3) $\lim _{x\rightarrow \infty }\dfrac {{x}^{2}-5}{{x}^{2}-1}=1$
步骤 1：将分子分母同时除以 ${x}^{2}$
$$\lim _{x\rightarrow \infty }\dfrac {{x}^{2}-5}{{x}^{2}-1} = \lim _{x\rightarrow \infty }\dfrac {1-\frac{5}{{x}^{2}}}{1-\frac{1}{{x}^{2}}}$$
步骤 2：当 $x\rightarrow \infty$ 时，$\frac{5}{{x}^{2}}\rightarrow 0$ 和 $\frac{1}{{x}^{2}}\rightarrow 0$
$$\lim _{x\rightarrow \infty }\dfrac {1-\frac{5}{{x}^{2}}}{1-\frac{1}{{x}^{2}}} = 1$$
(4) $\lim _{x\rightarrow {2}^{-}}\sqrt {4-{x}^{2}}=0$
步骤 1：当 $x\rightarrow {2}^{-}$ 时，$4-{x}^{2}\rightarrow 0$
$$\lim _{x\rightarrow {2}^{-}}\sqrt {4-{x}^{2}} = \sqrt{0} = 0$$
(5) $\lim _{x\rightarrow 0}\cos x=\cos 0$
步骤 1：直接代入 $x=0$
$$\lim _{x\rightarrow 0}\cos x = \cos 0 = 1$$