Direct Substitution

Direct substitution is the most basic and direct method for evaluating limits. When a function is continuous at the limit point, we can directly substitute the value of the limit point into the function to evaluate it.

Basic Principle

Basic principle of direct substitution

If the function f(x)f(x) is continuous at the point aa, then:

lim⁡x→af(x)=f(a)\lim_{x \to a} f(x) = f(a)

Applicable Conditions

1. The Function Is Continuous at the Limit Point

  • The function is defined at the point
  • The limit of the function at the point equals its value
  • The left-hand limit equals the right-hand limit at the point

2. Common Continuous Functions

  • Polynomial functions: P(x)=anxn+an−1xn−1+⋯+a1x+a0P(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0
  • Rational functions: R(x)=P(x)Q(x)R(x) = \frac{P(x)}{Q(x)} (where Q(a)≠0Q(a) \neq 0)
  • Exponential functions: exe^x
  • Logarithmic functions: ln⁡x\ln x (x>0x > 0)
  • Trigonometric functions: sin⁡x,cos⁡x,tan⁡x\sin x, \cos x, \tan x, etc.

Solving Steps

  1. Check whether the function is defined at the limit point
  2. Determine whether the function is continuous at the point
  3. Substitute directly to evaluate

Worked Examples

Example 1

Find the limit lim⁡x→3(2x2−5x+1)\lim_{x \to 3} (2x^2 - 5x + 1)

Reference Answer(1 个标签)
direct substitution

Idea: This is a polynomial function, which is continuous everywhere, so we can substitute directly.

Detailed steps:

  1. Check the type of function: f(x)=2x2−5x+1f(x) = 2x^2 - 5x + 1 is a polynomial function
  2. A polynomial function is continuous everywhere
  3. Substitute x=3x = 3 directly: f(3)=2(3)2−5(3)+1=2(9)−15+1=18−15+1=4f(3) = 2(3)^2 - 5(3) + 1 = 2(9) - 15 + 1 = 18 - 15 + 1 = 4

Answer: lim⁡x→3(2x2−5x+1)=4\lim_{x \to 3} (2x^2 - 5x + 1) = 4

Example 2

Find the limit lim⁡x→2x2+3x−2x−1\lim_{x \to 2} \frac{x^2 + 3x - 2}{x - 1}

Reference Answer(1 个标签)
direct substitution

Idea: This is a rational function, so we need to check whether the denominator is zero.

Detailed steps:

  1. Check the denominator: when x=2x = 2, the denominator x−1=2−1=1≠0x - 1 = 2 - 1 = 1 \neq 0
  2. A rational function is continuous wherever the denominator is nonzero
  3. Substitute x=2x = 2 directly: 22+3(2)−22−1=4+6−21=81=8\frac{2^2 + 3(2) - 2}{2 - 1} = \frac{4 + 6 - 2}{1} = \frac{8}{1} = 8

Answer: lim⁡x→2x2+3x−2x−1=8\lim_{x \to 2} \frac{x^2 + 3x - 2}{x - 1} = 8

Notes

1. When the Denominator Is Zero

If substitution makes the denominator zero, direct substitution cannot be used; other methods are needed.

2. When the Radicand Is Negative

If substitution makes the radicand negative, the function is undefined at the point.

3. Logarithmic Functions

The argument of a logarithmic function must be positive.

Practice Problems

Exercise 1

Find the limit lim⁡x→1(x3−2x2+x+5)\lim_{x \to 1} (x^3 - 2x^2 + x + 5)

Reference Answer(1 个标签)
direct substitution

Idea: A polynomial function, so substitute directly.

Detailed steps:

  1. f(x)=x3−2x2+x+5f(x) = x^3 - 2x^2 + x + 5 is a polynomial function
  2. A polynomial function is continuous everywhere
  3. Substitute x=1x = 1 directly: f(1)=13−2(1)2+1+5=1−2+1+5=5f(1) = 1^3 - 2(1)^2 + 1 + 5 = 1 - 2 + 1 + 5 = 5

Answer: lim⁡x→1(x3−2x2+x+5)=5\lim_{x \to 1} (x^3 - 2x^2 + x + 5) = 5

Exercise 2

Find the limit lim⁡x→0sin⁡xx\lim_{x \to 0} \frac{\sin x}{x}

Reference Answer(1 个标签)
direct substitution

Idea: This is an important limit; it cannot be evaluated by direct substitution, so we need other methods.

Detailed steps:

  1. When x=0x = 0, the numerator sin⁡0=0\sin 0 = 0 and the denominator x=0x = 0
  2. This is a 00\frac{0}{0} indeterminate form, so direct substitution cannot be used
  3. We need to use the important limit: lim⁡x→0sin⁡xx=1\lim_{x \to 0} \frac{\sin x}{x} = 1

Answer: lim⁡x→0sin⁡xx=1\lim_{x \to 0} \frac{\sin x}{x} = 1


Summary

Symbols Used in This Article

符号类型读音/说明在本文中的含义
lim⁡x→af(x)\lim_{x \to a} f(x)数学符号limit as x approaches aThe limit of the function as x→ax \to a
f(a)f(a)数学符号f of aThe value of the function at the point aa
aa数学符号aThe limit point
P(x)P(x)数学符号P of xA polynomial function
Q(x)Q(x)数学符号Q of xThe denominator of a rational function
exe^x数学符号e to the xThe exponential function
ln⁡x\ln x数学符号natural logarithm of xThe natural logarithmic function
sin⁡x,cos⁡x,tan⁡x\sin x, \cos x, \tan x数学符号sine/cosine/tangent of xTrigonometric functions
00\frac{0}{0}数学符号zero over zeroIndeterminate form

中英对照

中文术语英文术语音标说明
直接代入法direct substitution/dəˈrekt ˌsʌbstɪˈtjuːʃən/The method of directly substituting the limit point
连续函数continuous function/kənˈtɪnjuəs ˈfʌŋkʃən/A function with no breaks
多项式函数polynomial function/ˌpɒlɪˈnəʊmiəl ˈfʌŋkʃən/A function that is a polynomial
有理函数rational function/ˈræʃənəl ˈfʌŋkʃən/A function that is a polynomial divided by a polynomial
指数函数exponential function/ˌekspəˈnenʃəl ˈfʌŋkʃən/A function of the form axa^x
对数函数logarithmic function/ˌlɒɡəˈrɪðmɪk ˈfʌŋkʃən/A function involving logarithms
三角函数trigonometric function/ˌtrɪɡənəˈmetrɪk ˈfʌŋkʃən/Functions such as sine and cosine
不定式indeterminate form/ˌɪndɪˈtɜːmɪnət fɔːm/A limit form that cannot be determined directly