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Here the function $f(x)=1$ is plotted. As $x \to 0$, the $y$-values head to
1. Well, more to the point, they never vary from 1!
Quite a boring function.... |
Here the function $f(x)=x$ is plotted. As $x \to 0$, the $y$-values head to
0. And the closer $x$ gets to 0, the closer $y$ gets to 0. That's the way it's
supposed to work!
Also a rather boring function: so predictable! |
What can we do with those? We need more horsepower....
So what is this limit:
This sum (and many other operations) satisfy the same pattern:
So, in particular,
Suppose that $\lim_{x \to a} f(x)$ and $\lim_{x \to a} g(x)$ exist. Then \[ \lim_{x \to a} [f(x)+g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x) \] \[ \lim_{x \to a} [f(x)-g(x)] = \lim_{x \to a} f(x) - \lim_{x \to a} g(x) \] \[ \lim_{x \to a} [cf(x)] = c\lim_{x \to a} f(x) \] \[ \lim_{x \to a} [f(x)g(x)] = \lim_{x \to a} f(x) \lim_{x \to a} g(x) \] \[ \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim\limits_{x \to a} f(x)}{\lim\limits_{x \to a} g(x)} {\text{ if }} \lim_{x \to a} g(x) \ne 0 \]
\[ \lim\limits_{x\to{0}}\sqrt{x} \]
Because the function isn't even defined as x approaches 0 from the left (square roots of negative numbers are imaginary). So we have only a one-sided limit, that \[ \lim\limits_{x\to{0^+}}\sqrt{x}=0 \] Of course we can say that \[ \lim\limits_{x\to{1}}\sqrt{x}=1 \] Limits exists everywhere else on the domain of the sqrt function. |
This is the conclusion of Exercise #55, p. 71, which says that polynomials are continuous, which we'll consider more in section 1.8.
How do we know? Well an nth degree polynomial can be written as a sum of constants $c_i$ multiplying powers of $x$, $x^i$: \[ P(x)=\sum_{i=0}^n c_ix^i \] Then the limit of the sum is the sum of the limits: \[ \lim_{x\to a}P(x)=\lim_{x\to a}\sum_{i=0}^n c_ix^i =\sum_{i=0}^n \lim_{x\to a}(c_ix^i) \] and then we use the constant multiple rule and the power rules, and we're done: \[ \lim_{x\to a}P(x) =\sum_{i=0}^n c_i\lim_{x\to a}x^i =\sum_{i=0}^n c_ia^i = P(a) \]
Exercise #56 says it's also true for rational functions, for every element of their domain.
In this case, both show examples of a slightly different version of this most important definition, the derivative at a point $x=a$:
What's happening to the $y$-values?
Note: three things have to happen:
Otherwise $f$ is discontinuous at $a$.
There are various kinds of discontinuity (which we've already seen):
This function has a limit at zero (-.5), but is not defined there. If $f$ is not defined at $x=0$, then it cannot be continuous there. We can fix this, by the way.... Just define $f(0)$ to be the limit (the closed dot in the graph above).
If $g$ is continous at $x=c$ and $f$ is continous at $g(c)$ then
$F(x)$ is continuous at $x=c$.