Limits of Polynomial and Rational Functions:
1. If p is a polynomial function and c is a real number, then lim x-->c p(x) = p(c).
2. If r is a rational function given by r(x) = p(x) / q(x), and c is a real number such that q(c) does not equal 0, then lim x--> c r(x) = r(c) = p(c) / q(c), q(c) does not equal 0.
Methods of Evaluating Limits:
1. Direct substitution (plug-in).
2. Cancellation technique (factor/cancel).
3. Rationalization technique (multiply radicals by conjugate).
Once you reach the intermediate form (0/0), use either the cancellation or rationalization method to evaluate the limit. If it contains a radical, use the rationalization technique. Finally, plug the value of x
in the equation using direct substitution. This will give you the value of the limit.
Example 1: Evaluating Limits By Direct Substitution
Find each of the following limits.
a) lim x-->-1 (x^2 + x - 6)
b) lim x-->-1 (x^2 + x - 6 / x + 3)
The first function is a polynomial function and the second is a rational function (with a nonzero denominator at x = -1). Thus, you can evaluate the limits by direct substitution.
a) lim x-->-1 (x^2 + x - 6) = (-1)^2 + (-1) - 6 = -6
b) lim x-->-1 (x^2 + x - 6) / x + 3 = -6 / -1 + 3 = -3
Example 2: Evaluating Limits By Cancellation
Find the limit: lim x--> -3 (x^2 + x - 6) / x + 3.
Begin by factoring the numerator and canceling any common factors.
lim x--> -3 (x^2 + x - 6) / x + 3 (Factor numerator)
lim x--> -3 (x - 2)(x + 3) / x + 3 (Cancel common factor)
lim x--> -3 (x - 2) (Simplify)
lim = 5 (Direct substitution)
Find the limit: lim x--> 1 (x - 1) / x^3 - x^2 + x - 1
Begin by substituting x = 1 into the numerator and denominators.
1 - 1 = 0 (Numerator is 0 when x = 1)
(1)^3 - (1)^2 + 1 - 1 = 0 (Denominator is 0 when x = 1)
Because both the numerator and denominator are zero when x = 1, direct substitution will not yield the limit. To fins the limit, you should factor the numerator and denominator, cancel any common factors, and then try direct substitution again.
lim x--> 1 (x - 1) / x^3 - x^2 + x - 1 =
lim x--> 1 (x - 1) / (x - 1)(x^2 + 1) (Factor denominator)
lim x--> 1 (x - 1) / (x - 1)(x^2 + 1) (Cancel common factor)
lim x--> 1 (1) / (x^2 + 1) (Simplify)
(1) / (1)^2 + 1 (Substitute)
lim = 1 / 2 (Simplify)
Example 3: Evaluating Limits By Rationalization
Find the limit: (x + 1)^1/2 - 1 / x
By direct substitution, you obtain the indeterminate form 0 / 0.
lim x-->0 (x + 1)^1/2 - 1 / x = 0 / 0
In this casem you can rewrite the fraction by rationalizing the numerator.
(x + 1)^1/2 - 1 / x = ((x + 1)^1/2 - 1 / x)((x + 1)^1/2 - 1 / (x + 1)^1/2 + 1))
= (x + 1) - 1 / x((x + 1)^1/2 + 1)) (Multiply)
= x / x((x + 1)^1/2 + 1)) (Simplify)
= x / x((x + 1)^1/2 + 1)) (Cancel common factor)
= 1 / (x + 1)^1/2 + 1) (Simplify)
Now you can evaluate the limit by direct substitution.
lim x-->0 (x + 1) ^1/2 - 1 / x
lim x--> 0 1 / (x + 1)^1/2 + 1
= 1 / 1 + 1
= 1 / 2
You can reinforce your conclusion that the limit is 1/2 by constructing a table, or by sketching a graph.
One-Sided Limits
lim x--> c- f(x) = L (Limit from the left)
lim x--> c+ f(x) = L (Limit from the right)
The function approaches a different value from the left side of c than it approaches from the right of c.
Thus, a limit can fail to exist. This is a concept known as a one-sided limit.
Example 4: Evaluating One-Sided Limits
Find the limit as x--> 0 from the left and the limit as x--> 0 from the right for the function given by:
f(x) = |2x| / x.
From the graph of f, shown below, you can see that f(x) = -2 for all x < 0. Therefore, the limit frol the left is lim x--> 0 |2x| / x = -2 (Limit from the left). Because f(x) = 2 for all x > 0, the limit from the right is lim x--> 0 |2x| / x = 2 (Limit from the right).
Existence of a Limit:
If f is a function and c and L are real numbers, then lim x--> c f(x) = L if and only if both the left and right limits are equal to L.
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