IBDP. AA. HL. Differentiation. Limits

Opening Problem

We will define xn as 0.333333… , with n 3s after the decimal point (e.g. so x2 = 0.33). n is restricted to positive integers.

nxn3xn1 – 3xn
1
2
3
4
5
10
  1. Copy and complete the table above. In the last column, how many zeroes are there between the decimal point and the 1?
  2. Consider x100, which contains 100 3s. In the number 1 – 3x100, how many zeroes are there between the decimal point and the 1?
  3. As n gets progressively larger, how many zeroes are there in 1 – 3xn between the decimal point and the 1? How close does xn get to 1/3?

The opening problem shows us an example of a limiting process. As n tends to infinity the sequence approaches the limiting value 1/3. Another way of saying this is that for any number we might choose which is close to 1/3, we can find a value of n for which the sequence is closer and remains closer. We write \lim_{ x \to \infty} = \frac{1}{3} .

We will use the following definition of a limit, which is sufficiently accurate for this course:

Definition of Limit: If f(x) is as close as we like to some real number A for all x sufficiently close to (but not equal to) a, then we say that f(x) has a limit of A as x approaches a, and we write \lim_{ x \to a} f(x) = A .

The function may or may not be defined at x=a and this is irrelevant to the definition of the limit of f as x approaches a. We are interested in the behaviour of the function as x gets very close to a.

For instance, consider f(x) = \frac{5x+x^2}{x} as x \to 0 . If we substitute 0 in directly we get the meaningless value 0/0. However, if we rewrite the function as f(x) = \frac{x(5+x)}{x} we can see that for values close to zero f(x) = 5+x.

The graph of y=f(x) is the straight line y = x+5 with a hole, or a discontinuity, at (0,5). Even though f(0) doesn’t exist, the limit still exists, so f(x) \to 5 as x \to 0 . we write \lim_{ x \to 0} \frac{5x+x^2}{x} = 5 .

Limit Laws

For functions f(x) and g(x) with \lim_{ x \to a} f(x) = l and \lim_{ x \to a} g(x) = m , where a, l and m are real numbers, the following is true:

  • \lim_{ x \to a} (f(x) \pm g(x) = l \pm m
  • \lim_{ x \to a} f(x)g(x) = lm
  • \lim_{ x \to a} ( \frac{f(x)}{g(x)}) = \frac{l}{m}, provided m \neq 0

Worked Example

Evaluate the following:

  • \lim_{ x \to 2} x^2
  • \lim_{ x \to 0} /frac{x^2 + 3x}{x}
  • \lim_{ x \to 3} /frac{x^2 - 9}{x-3}

Exercise

Answers