This is a comprehensive review of limits of functions.
Main Types of Indeterminate Forms
- 0/0：factoring／rationalization／trigonometric and exponential formulas
- ∞/∞：divide by the highest-degree term
- ∞−∞：rationalization／factoring out
- 0·∞：rewrite as a fraction to obtain 0/0 or ∞/∞
- 1^∞：rewrite in the form (1+a/x)^x
Example ①
lim_{x→0} (sin x − tan x)/x³
Solution:sin x − tan x = sin x·(cos x − 1)/cos x。
={sin x/x}·{(cos x−1)/x²}·(1/cos x) → 1·(−1/2)·1 = −1/2
lim_{x→0} (sin x − tan x)/x³
Solution:sin x − tan x = sin x·(cos x − 1)/cos x。
={sin x/x}·{(cos x−1)/x²}·(1/cos x) → 1·(−1/2)·1 = −1/2
Determining Coefficients
If the denominator approaches 0 and the limit is finite, the numerator must also approach 0.
Example ②
Find a and b such that lim_{x→2} (x²+ax+b)/(x−2) = 5。
Solution:Numerator→0：4+2a+b=0。Numerator=(x−2)(x+c), limit=2+c=5, c=3。
x²+ax+b=(x−2)(x+3)=x²+x−6, so a=1、b=−6
Find a and b such that lim_{x→2} (x²+ax+b)/(x−2) = 5。
Solution:Numerator→0：4+2a+b=0。Numerator=(x−2)(x+c), limit=2+c=5, c=3。
x²+ax+b=(x−2)(x+3)=x²+x−6, so a=1、b=−6
Squeeze Theorem
For lim_{x→0} x²·sin(1/x), since 0 ≦ |x²·sin(1/x)| ≦ x² → 0, the limit is 0。
Key Points
- Identify the type of indeterminate form
- Determining coefficients: if the denominator→0 and the limit is finite, the numerator→0
- The squeeze theorem is effective when the absolute value is bounded
- Factor out constant multiples and create the form of a known formula
Practice Problems
- lim_{x→0} (√(1+x)−1)/x
- lim_{x→1} (x²+ax−a−1)/(x−1)
- Use the squeeze theorem to find lim_{x→0} x·sin(1/x)
Answers and Explanations
- Answer:1/2
Rationalize. 1/(√(1+x)+1) → 1/2。 - Answer:2+a
(x−1)(x+1+a)/(x−1) → 1+1+a = 2+a。 - Answer:0
|x·sin(1/x)| ≦ |x| → 0。