CHANGE AND LIMITS 07
Limit Laws
Limit laws let us calculate many limits from simpler limits.
The main idea
Suppose
\[ \lim_{x\to a}f(x)=L \]and
\[ \lim_{x\to a}g(x)=M. \]Then limits usually behave in the same way as ordinary algebraic operations.
1. Constant rule
The limit of a constant is the constant itself:
For example,
\[ \lim_{x\to3}7=7. \]2. Sum rule
The limit of a sum is the sum of the limits:
\[ \lim_{x\to a}[f(x)+g(x)] = L+M. \]For example,
\[ \lim_{x\to2}(x^2+x) = 4+2 = 6. \]3. Difference rule
The limit of a difference is the difference of the limits:
\[ \lim_{x\to a}[f(x)-g(x)] = L-M. \]4. Constant multiple rule
A constant multiplier can be taken outside the limit:
\[ \lim_{x\to a}[cf(x)] = cL. \]For example,
\[ \lim_{x\to2}3x^2 = 3(4) = 12. \]5. Product rule
The limit of a product is the product of the limits:
For example,
\[ \lim_{x\to2}x(x+1) = 2(3) = 6. \]6. Quotient rule
The limit of a quotient is the quotient of the limits, provided the denominator does not approach zero:
\[ \lim_{x\to a} \frac{f(x)}{g(x)} = \frac{L}{M}, \qquad M\neq0. \]For example,
\[ \lim_{x\to2} \frac{x+1}{x+3} = \frac35. \]7. Power rule
For a positive integer \(n\),
\[ \lim_{x\to a}[f(x)]^n = L^n. \]For example,
\[ \lim_{x\to3}x^2 = 3^2 = 9. \]8. Direct substitution
For many familiar functions, the limit can be found simply by substituting the approaching value.
For example,
\[ \lim_{x\to2}(x^2+3x-1) \]is
\[ 2^2+3(2)-1 = 9. \]This works because polynomial functions are continuous.
9. When substitution fails
Consider
\[ \lim_{x\to1} \frac{x^2-1}{x-1}. \]Direct substitution gives
\[ \frac00, \]which is undefined.
This does not mean the limit does not exist. It means the expression must be simplified first.
Factorising,
\[ x^2-1=(x-1)(x+1). \]So, for \(x\neq1\),
\[ \frac{x^2-1}{x-1}=x+1. \]Therefore,
\[ \lim_{x\to1} \frac{x^2-1}{x-1} = \lim_{x\to1}(x+1) = 2. \]10. Why \(0/0\) is important
An expression such as
\[ \frac00 \]is called an indeterminate form.
It does not determine the limit by itself. Different functions can produce the form \(0/0\) but have different limits.
So when substitution gives \(0/0\), we need more analysis.
11. Example using several laws
Find
\[ \lim_{x\to2} \frac{x^2+3x+1}{x+4}. \]Using the limit laws,
\[ \lim_{x\to2}(x^2+3x+1) = 4+6+1 = 11, \]and
\[ \lim_{x\to2}(x+4)=6. \]Therefore,
\[ \lim_{x\to2} \frac{x^2+3x+1}{x+4} = \frac{11}{6}. \]12. The practical idea
When finding a limit, first try direct substitution.
If substitution gives an ordinary number, the calculation is usually finished. If it gives an indeterminate form such as \(0/0\), further work is needed.