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By the way, the limit is a hungry, hungry operator: it has to eat something -- it's not okay to leave it floating out there alone.
$\lim_{x\to{a}}$
doesn't make any sense: it must eat something:
$\lim_{x\to{a}}f(x)$
or, better yet,
$\lim_{x\to{a}}\left(f(x)\right)$
The limit is gobbling up that $f(x)$!
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,
\[ \lim_{x\to{0}}\sqrt{x} \]
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 \[ \lim_{x\to{0^+}}\sqrt{x}=0 \] Of course we can say that \[ \lim_{x\to{1}}\sqrt{x}=1 \] 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.
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 is discontinuous at
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 $g$ is continous at $x=c$ and $f$ is continous at $g(c)$ then
$F(x)$
is continuous at .