Welcome to Calculus Revisited! This is blog entry #4 of a 21 blog entry series. Today, we will cover the basics of limits.
Limit: The limit of a function f(x) is a value of which a function approaches as x creeps closer and closer to it's target x=a. This is not the same as f(a).
In notation:
lim f(x) = L
x→a
The limit may or may not exist.
Left Side Limit: The limit of a function f(x) as x approaches a from the left side. Hence, x < a.
In notation:
lim f(x) = L
x→ a-
Right Side Limit: The limit of a function f(x) as x approaches a from the right side. Hence, x > a.
In notation:
lim f(x) = L
x→ a+
If the left side limit is equal to the right side limit, then the general limit exists at L.
Continuous Function: A function f(x) is continuous if a limit exists for each x in the domain (or specified interval), and that limit is the same as the function's value. It has been often said that you can graph continuous functions without lifting a pencil.
Common continuous functions include:
f(x) = p(x) (polynomials a_n * x^n + ... + a_0)
f(x) = e^x
f(x) = ln x (for x>0 only)
f(x) = sin x
f(x) = cos x
Properties of Limits
lim (f(x) + g(x)) = lim f(x) + lim g(x)
lim c *f(x) = c * lim f(x), c is a constant
Common Ways to Attack Limits
1. If f(x) is continuous, then
lim f(x) = L
x →a
for all x.
2. "Calculator Method": plug in various x_i as x approaches closer and closer to x = a (but not at x = a), observe the results and make a educated conclusion.
3. Graph the function. If graphing calculators are allowed, this is the time to use them.
Problems
1. Find
lim x^2 + 1
x→2
Using the "calculator method":
Left Side Limit: (x approaches 2 with x < 2)
f(2 - .01) = 4.9601
f(2 - .001) = 4.99601
f(2 - .0001) = 4.99960001
f(2 - 10^-9) = 4.999999996
The value is getting close to 5.
Right Side Limit: (x approaches 2 with x > 2)
f(2 + .01) = 5.0401
f(2 + .001) = 5.004001
f(2 + .0001) = 5.00040001
f(2 + 10^-9) = 5.000000004
The value is getting close to 5.
We can reasonably conclude that
lim x^2 + 1 = 5 as x → 2
We could have also observed that x^2 + 1 is continuous everywhere and figured the limit out by plugging in 2 for x.
Now let's go to a case where f(x) is not continuous everywhere.
2. Find
lim 1/(x-2)
x→2
f(2) = 1/0. So plugging in x=2 does not work here.
Left Side Limit: (x approaches 2 with x < 2)
f(2 - .001) = -1,000
f(2 - .0001) = -10,000
f(2 - .00001) = -100,000
f(2 - 10^-9) = -1,000,000,000
Note that f(x) is getting to be a very large negative number, towards -∞
Right Side Limit: (x approaches 2 with x > 2)
f(2 + .001) = 1,000
f(2 + .0001) = 10,000
f(2 + .00001) = 100,000
f(2 + 10^-9) = 1,000,000,000
Note that f(x) is getting to be a very large positive number, towards ∞
But -∞ ≠ ∞
3. Find
lim (sin x)/x
x→ 0
Again, f(0) = 0/0, the plugging it won't work.
Left Side Limit:
f(-.001) = .9999998333
f(-.0001) = .9999999983
f(-.00001) = 1 (calculator returns 1)
Right Side Limit:
f(.001) = .9999998333
f(.0001) = .9999999983
f(.00001) = 1 (calculator returns 1)
From the "Calculator Method":
lim (sin x)/x = 1
x → 0
The next time we will working with derivatives. See you next time, Eddie.
This blog is property of Edward Shore. © 2012
