Thursday, April 26, 2012

Calculus Revisited #4: Limits

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

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