9
$$\eqalign{
& \mathop {\lim }\limits_{x \to 4} \frac{{{x^2} - 6x + 8}}{{{x^2} - 5x + 4}} \cr
& = \mathop {\lim }\limits_{x \to 4} \frac{{\left( {x - 4} \right)\left( {x - 2} \right)}}{{\left( {x - 4} \right)\left( {x - 1} \right)}} \cr
& = \mathop {\lim }\limits_{x \to 4} \frac{{x - 2}}{{x - 1}} = \frac{2}{3} \cr} $$
10
$$\mathop {\lim }\limits_{n \to \infty } (1 + \frac{1}{2} + \frac{1}{4} + ... + \frac{1}{{{2^n}}})$$
利用等比数列求和公式，
$$\eqalign{
& 1 + \frac{1}{2} + \frac{1}{4} + ... + \frac{1}{{{2^n}}} \cr
& = \frac{{1 - {{(\frac{1}{2})}^n}}}{{1 - \frac{1}{2}}} = 2 - \frac{2}{{{2^n}}} \cr} $$
$$\eqalign{
& \mathop {\lim }\limits_{n \to \infty } (1 + \frac{1}{2} + \frac{1}{4} + ... + \frac{1}{{{2^n}}}) \cr
& = \mathop {\lim }\limits_{n \to \infty } (2 - \frac{2}{{{2^n}}}) = 2 \cr} $$
11
$$\mathop {\lim }\limits_{n \to \infty } \frac{{1 + 2 + 3 + ... + (n - 1)}}{{{n^2}}}$$
由等差数列前n项和公式
$$\eqalign{
& 1 + 2 + 3 + ... + (n - 1) \cr
& = (n - 1) + \frac{{(n - 1)(n - 2)}}{2} \cr
& = \frac{{n(n - 1)}}{2} \cr
& \mathop {\lim }\limits_{n \to \infty } \frac{{1 + 2 + 3 + ... + (n - 1)}}{{{n^2}}} \cr
& = \mathop {\lim }\limits_{n \to \infty } \frac{{n(n - 1)}}{{2{n^2}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{n - 1}}{{2n}} \cr
& = \mathop {\lim }\limits_{n \to \infty } (\frac{1}{2} - \frac{1}{{2n}}) = \frac{1}{2} \cr} $$
12
$$\eqalign{
& \mathop {\lim }\limits_{x \to 1} (\frac{1}{{1 - x}} - \frac{3}{{1 - {x^3}}}) \cr
& = \mathop {\lim }\limits_{x \to 1} \frac{{1 + x + {x^2} - 3}}{{(1 - x)(1 + x + {x^2})}} \cr
& = \mathop {\lim }\limits_{x \to 1} \frac{{(x - 1)(x + 2)}}{{(1 - x)(1 + x + {x^2})}} \cr
& = \mathop {\lim }\limits_{x \to 1} \frac{{ - (1 - x)(x + 2)}}{{(1 - x)(1 + x + {x^2})}} \cr
& = \mathop {\lim }\limits_{x \to 1} \frac{{ - (x + 2)}}{{1 + x + {x^2}}} \cr
& = \mathop {\lim }\limits_{x \to 1} \frac{{ - x - 2}}{{1 + x + {x^2}}} = - 1 \cr} $$
13
$$\eqalign{
& \mathop {\lim }\limits_{x \to 2} \frac{{{x^3} + 2{x^2}}}{{{{\left( {x - 2} \right)}^2}}} \cr
& = \mathop {\lim }\limits_{x \to 2} \frac{1}{{{{\left( {x - 2} \right)}^2}}}\mathop {\lim }\limits_{x \to 2} \left( {{x^3} + 2{x^2}} \right) \cr
& = 16\mathop {\lim }\limits_{x \to 2} \frac{1}{{{{\left( {x - 2} \right)}^2}}} = \infty \cr} $$
14
$$\eqalign{
& \mathop {\lim }\limits_{x \to \infty } \frac{{{x^2}}}{{2x + 1}} \cr
& = \mathop {\lim }\limits_{x \to \infty } \frac{{x\frac{{{x^2}}}{x}}}{{x\frac{1}{x}\left( {2x + 1} \right)}} \cr
& = \mathop {\lim }\limits_{x \to \infty } \frac{x}{{2 + \frac{1}{x}}} = \infty \cr} $$
15
$$\eqalign{
& \mathop {\lim }\limits_{x \to \infty } \left( {2{x^3} - x + 1} \right) \cr
& = \mathop {\lim }\limits_{x \to \infty } 1 + \mathop {\lim }\limits_{x \to \infty } \left( {2{x^3} - x} \right) \cr
& = 1 + \mathop {\lim }\limits_{x \to \infty } \left( {2{x^3} - x} \right) \cr
& = \mathop {\lim }\limits_{x \to \infty } {x^3}\left( {2 - \frac{1}{{{x^2}}}} \right) + 1 = \infty \cr} $$
16
$$\eqalign{
& \mathop {\lim }\limits_{x \to 0} {x^2}\sin \frac{1}{x} \cr
& = \mathop {\lim }\limits_{x \to 0} {x^2}\mathop {\lim }\limits_{x \to 0} \sin \frac{1}{x} = 0 \cr} $$
备注：无穷小与有界函数的乘积是无穷小
17
$$\eqalign{
& \mathop {\lim }\limits_{x \to \infty } \frac{1}{x}\arctan x \cr
& = \mathop {\lim }\limits_{x \to \infty } \frac{1}{x}\mathop {\lim }\limits_{x \to \infty } \arctan x \cr
& = \frac{\pi }{2}\mathop {\lim }\limits_{x \to \infty } \frac{1}{x} = 0 \cr} $$
18
$$\eqalign{
& \mathop {\lim }\limits_{x \to 0} \frac{{\sin \omega x}}{x} \cr
& = \mathop {\lim }\limits_{x \to 0} \omega \frac{{\sin \omega x}}{{\omega x}} \cr
& = \omega \mathop {\lim }\limits_{x \to 0} \frac{{\sin \omega x}}{{\omega x}} \cr
& = \omega \cr} $$
19
$$\eqalign{
& \mathop {\lim }\limits_{x \to 0} \frac{{\tan 3x}}{x} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{\sin 3x}}{{\cos 3x}}}}{x} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{\sin 3x}}{{x\cos 3x}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{1}{{\cos 3x}}\mathop {\lim }\limits_{x \to 0} \frac{{\sin 3x}}{x} \cr
& = 3\mathop {\lim }\limits_{x \to 0} \frac{{\sin 3x}}{{3x}} \cr
& = 3 \cr} $$
20
$$\eqalign{
& \mathop {\lim }\limits_{x \to 0} \frac{{\sin 2x}}{{\sin 5x}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{\sin 2x}}{{2x}}}}{{\frac{{\sin 5x}}{{2x}}}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{1}{5}\frac{{\sin 2x}}{{2x}}}}{{\frac{1}{2}\frac{{\sin 5x}}{{5x}}}} \cr
& = \frac{{\frac{1}{5}\mathop {\lim }\limits_{x \to 0} \frac{{\sin 2x}}{{2x}}}}{{\frac{1}{2}\mathop {\lim }\limits_{x \to 0} \frac{{\sin 5x}}{{5x}}}} \cr
& = \frac{2}{5} \cr} $$
