We have arrived at the 15th blog entry of the 21 entry series of Calculus Revisited. It is all about infinity today, and some quick tips with deal with infinity and limits.
Infinite Limits
General Rules for Polynomials
Let p(x) be a polynomial with degree n. The lead term of p(x) is a_n*x^n.
If n is even and a_n is positive:
* p(x) →+∞ as x → +∞
* p(x) → +∞ as x → -∞
If n is even and a_n is negative:
* p(x) → -∞ as x → +∞
* p(x) → -∞ as x → -∞
If n is odd and a_n is positive:
* p(x) → +∞ as x → +∞
* p(x) → -∞ as x → -∞
If n is odd and a_n is negative:
* p(x) → -∞ as x → +∞
* p(x) → +∞ as x → -∞
Generally: 1 / p(x) → 0 as x → +∞ or x → -∞
If p(x)/q(x) have the same degree, then
p(x)/q(x) → (lead coefficient of p(x)) / (lead coefficient of q(x))
as x → +∞ or x → -∞
General Rules for Other Functions
ln x → -∞ as x → -∞
ln x → +∞ as x → +∞
e^x → 0 as x → -∞
e^x → +∞ as x → +∞
e^-x → +∞ as x → -∞
e^-x → 0 as x → +∞
sin x has no limit as x → +∞ or x → -∞
cos x has no limit as x → +∞ or x → -∞
tan x has no limit as x → +∞ or x → -∞
Note that sin x and cos x are both bounded between -1 and 1.
L'Hospital's Rule
For indeterminate forms 0/0, +∞/+∞, -∞/-∞ you can use the rule:
lim f(x)/g(x) =
x → a
lim f'(x)/g'(x)
x → a
Let's go over some problems.
Problems
1. Find: lim (1/x * sin) as x → ∞
We know that:
lim 1/x = 0 as x → ∞ and sin x is bounded.
Then the overall limit is 0 * (a number between -1 and 1) = 0
2. Find lim (e^(-x^2)) as x → ∞
In general, e^-x → 0 as x → ∞
Even though x^2 grows, e^-x shrinks faster and is dominant.
Hence lim (e^(-x^2)) = 0 as x → ∞.
The next three problems uses the p(x)/q(x) rules.
3. lim (x^2 + 1)/(x - 2) as x → ∞
Degree of numerator = 2
Degree of denominator = 1
Numerator will grow faster than the denominator.
Then lim (x^2 + 1)/(x - 2) → ∞ as x → ∞
4. lim (x^2 + 1)/(x^2 - 1) as x → ∞
Degree of numerator = 2
Degree of denominator = 2
Then lim (x^2 + 1)/(x^2 - 1) = 1/1 = 1 as x → ∞
5. lim (x^2 + 1)/(x^3 - 1) as x → ∞
Degree of numerator = 2
Degree of denominator = 3
Denominator will faster than numerator.
Then lim (x^2 + 1)/(x^3 - 1) = 0 as x → ∞
The next two problems uses L'Hospital's Rule.
6. lim (x^2 / e^x) as x → ∞
lim (x^2 / e^x) as → ∞ = ∞/∞ (Indeterminate form)
Applying the rule:
lim (2x / e^x ) as x → ∞ = ∞/∞ (Indeterminate form)
One more time:
lim (2 / e^x) as x → ∞ = 0
Therefore, lim (x^2 / e^x) = 0 as x → ∞
7. lim (sin x / x) as x → 0
lim (sin x / x) as x → 0 = 0/0 (Indeterminate form)
Applying the rule:
lim (cos x/1) as x→ 0 = 1/1 = 1
Therefore, lim (sin x / x) = 1 as x → 0
This concludes today's session. Next time we work with improper integrals (yes, there are such things!).
Have a great day,
Eddie
This blog is property of Edward Shore. © 2012
