Limits (Sequences, functions, graphs)

What do the numbers 1/1, 1/2, 1/3, 1/4, 1/5, . . . get closer and closer to? It may be clear that they approach zero, so we say the limit is 0. The nth term is 1/n, so the notation is lim (n->oo) 1/n = 0. (The -> means "approaches"; the oo is a cheesy infinity symbol.)

The same idea is to consider the function f(x) = 1/x . Then as x -> oo , f(x) -> 0 , so again we say lim (x->oo) f(x) = lim (x->oo) 1/x = 0. On the graph we'd have a horizontal asymptote at y = 0 since the output values approach 0 as the graph goes off to the right. (Click here to review functions or graphing.)

Now what if x -> a and we see if f(x) approaches any specific value L. If it does, then we say that lim (x->a) f(x) = L. By definition x->a means x is near a but not equal to a.

Example 1: Let f(x) = x + 3 . Then as x -> 3, f(x) -> 3 + 3, so lim (x->3) [x+3] = 6.

Example 2: Let g(x) = (x^2 - 9) / (x - 3) . By algebra, we have

g(x) = (x + 3)(x - 3) / (x - 3) and if x =/= 3 then we can cancel, so

g(x) = x + 3 if x =/= 3. Notice that f(3) = 6 but g(3) is undefined.

The f(x) from example 1 has domain "all real numbers," but the g(x) from example 2 has domain "all reals except 3," so they're different functions. But the limit as x -> 3 is the same in both cases: lim (x->3) f(x) = lim (x->3) g(x) = 6. Even though g(3) is undefined (it'd be 0 / 0) g(x) still has a limit (of 6), since x -> 3 implies x =/= 3.

Example 3: Some other interesting limits:

lim (x->0) [(sin x) / x] = 1

lim (x->0) [(1 + x)^(1/x)] = e = 2.71828 approx

lim (n->oo) [F(n+1) / F(n)] = (1 + Sqrt(5)) / 2 = 1.61805 approx.

The F(n) is the nth Fibonacci number of the sequence 1, 1, 2, 3, 5, 8, 13, . . .

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Differential Calculus

But on a parabola, like y = x^2, the direction keeps changing, so we'd expect that the "slope" doesn't stay constant. But how do you figure out the slope of a curve?

Let's take (3,9) as an example; if we join up (3,9) and a nearby point on the curve, say (3.1,9.61), and do "rise/run": (draw yourself a picture)

. slope = (y2 - y1) / (x2 - x1) = (9.61 - 9) / (3.1 - 3) = 0.61 / 0.1 = 6.1

If we take a closer point (3.02,3.02^2) = (3.02,9.1204) then the slope is

. slope = (9.1204 - 9) / (3.02 - 3) = 0.1204 / 0.02 = 6.02

In fact for any h, the slope between (3,9) and (3+h,(3+h)^2) will be

. slope = ((3+h)^2 - 9) / (3+h - 3) = (6h + h^2) / h = 6 + h

The line joining two points on a curve is called a secant line. The slope is msec.

As h gets smaller and the points get closer together, the slope of the secant line approaches 6. The line it approaches is called the tangent line. The slope is mtan.

. slope = msec = ((x+h)^2 - x^2) / (x+h - x) = (2xh + h^2) / h = 2x + h.

As h goes to 0, this becomes mtan = 2x. The expression for "the slope at any x" is a new function, derived from f(x), called the derivative of f, and denoted f'(x).

Here we have: if f(x) = x^2 then f'(x) = 2x.

Checking this for our old point (3,9) we see if x = 3 then f'(x) = 2x = 2(3) = 6. Ok!

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Related Rates

You guessed it -- define some variables, try to relate their rates (of change); we get a relation between their derivatives; technically known as a "differential equation."

Example: Spoze you blow 30 cu.in. of air per sec into a spherical balloon. How fast is the radius increasing when the balloon is 6 inches in diameter?

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Max/Min Problems

The philosophy here is to optimize some quantity Q that can depend on more than one thing (variable). Find a relation (constraint) and use it to get the Q in terms of one variable; then set the first derivative equal to zero and solve!

Example: Of all the rectangles of area A, which has the shortest diagonals?

Approach: Ok, use x = width, y = height; then xy = A; we want to minimize D = (x^2 + y^2) ;
we can sub in y = A/x since A is constant. Also we can minimize D^2 and D will be smallest.
D^2 = x^2 + (A/x)^2 ; now it's in terms of one variable; take the deriv. w.r.t. x :
(D^2) = (x^2 + (A/x)^2) = (x^2 + (A^2) x^(-2))
. . . . . . . . . . = 2x + A^2 (-2 x^(-3)) = 2x - (2 A^2)/(x^3) =set= 0 ;
2x = 2 A^2 / x^3 ; x^4 = A^2 ; x = -/A ; y = A/x = -/A ;
Analysis: The rectangle's a square, Captain Kirk.

Well, that's it for now. Check back often for new stuff!
Click below for other topics, or visit the ask dan page!
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