Calculus/Formal limits

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[edit] Formal definition of limit (optional)

In preliminary calculus, the definition of a limit is probably the most difficult concept to grasp. If nothing else, it took some of the most brilliant mathematicians 150 years to arrive at it.

The intuitive definition of a limit is adequate in most cases, as the limit of a function is the function of the limit. But what is our meaning of "close"? How close is close? We consider the following limit of a function:

\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac {\sin(x)} {x}.

You will have noted that f(0) = 0/0, and this is undefined. But the limit does exist and it equals 1. But how do we convert that intuition into mathematical sense?

We call L the limit of f(x) as x approaches c if for every positive number ε, there exists a number δ such that

\left| f(x) - L \right| < \epsilon

whenever

0 < \left| x - c \right| < \delta.

Understanding the parallels between the two definitions above is important: instead of saying ' f(x) approximately equals L ', the formal definition says that 'the difference between f(x) and L is less than any number epsilon'.


\lim_{x \to c} f(x) = L
if and only if
\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = L

\lim_{x \to c^-} f(x) means as x approaches c from the left, and \lim_{x \to c^+} f(x) as x approaches c from the right.


See how to make formal proofs.

Notice that we have not yet defined ' x approaches c ' yet; we will discuss this later.

Limits are often written

\lim_{x \rightarrow c} f(x)

which may be read "the limit of f(x) as x approaches c".

[edit] Examples

1) What is the limit of f(x) = x + 7 as x approaches 4?

There are two steps to answering such a question; first we must determine the answer -- this is where intuition and guessing is useful, as well as the informal definition of a limit. Then, we must prove that the answer is right. For this problem, the answer happens to be 11. Now, we must prove it using the definition of a limit:

Informal: 11 is the limit because when x is roughly equal to 4, f(x) = x + 7 approximately equals 4 + 7, which equals 11.

Formal: We need to prove that no matter what value of ε is given to us, we can find a value of δ such that

\left| f(x) - 11 \right| < \epsilon

whenever

\left| x - 4 \right| < \delta.

For this particular problem, letting δ equal ε works (see choosing delta for help in determining the value of delta to use). Now, we have to prove

\left| f(x) - 11 \right| < \epsilon

given that

\left| x - 4 \right| < \delta = \epsilon.

Since |x - 4| < ε, we know
|f(x) - 11|  = |x + 7 - 11| = |x - 4| < ε, which is what we wished to prove.


2) What is the limit of f(x) = x² as x approaches 4?

Formal: Again, we pull two things out of thin air; the limit is 16 (use the informal definition to find the limit of f(x)), and δ equals √(ε+16) - 4. Note that δ is always positive for positive ε. Now, we have to prove
\left| x^2 - 16 \right| < \epsilon
given that
\left| x - 4 \right| < \delta = \sqrt{\epsilon + 16} - 4.

We know that |x + 4| = |(x - 4) + 8| ≤ |x - 4| + 8  < δ + 8 (because of the triangle inequality), thus

\begin{matrix}
\left| x^2 - 16 \right| & = & \left| x - 4 \right| \cdot \left| x + 4 \right| \\  \\
\ & < & (\delta) \cdot (\delta + 8) \\  \\
\ & = & (\sqrt{16 + \epsilon} - 4) \cdot (\sqrt{16 + \epsilon} + 4) \\  \\
\ & = & (\sqrt{16 + \epsilon})^2 - 4^2 \\  \\
\ & = & \epsilon \end{matrix}.

3) Show that the limit of sin(1/x) as x approaches 0 does not exist.

Suppose the limit exists and is l. We will proceed by contradiction. Assume that l\neq 1, the case for l = 1 is similar. Choose ε = l - 1, then for every δ > 0, there exists a large enough n such that  0 < x_0 = \frac{1}{\pi /2 + 2\pi n} < \delta, but | sin(1 / x0) - l | = | 1 - l | = ε a contradiction.

The function sin(1/x) is known as the topologist's curve.

4) What is the limit of xsin(1 / x) as x approaches 0?

It is 0. For every ε > 0, choose δ = ε so that for all x, if 0 < | x | < δ, then | xsinx - 0 | < = | x | < ε as required.


5) Prove that the limit of 1/x as x approaches 0 does not exist?


6) Prove that the limit of f(x)={x, when x is rational; 0, when x is irrational} as x approaches c for ALL (c not equal to zero, c is real number) does not exist?

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