For each x in X, there exists the unique y in Y
Therefore, the limit of a function is about the relation between x's and y's.
The slope of secant line of \( (x_0,f(x_0))\) and \((x_0+\Delta x,f(x_0+\Delta x))\) is given by
\[ \frac{\Delta y}{\Delta x}=\frac{f(x_0 + \Delta x) - f(x_0)}{x_0+\Delta x-x_0}=\frac{f(\Delta x+x_0)-f(x_0)}{\Delta x} \]
For example, if \(\Delta x=1\), then
\[ \frac{\Delta y}{\Delta x}=\frac{f(x_0+\Delta x)-f(x_0)}{\Delta x} \]
\[ =\frac{f(2+1)-f(2)}{1}=\frac{2.25-1}{1}=1.25 \]
\( \Delta x \) | 1 | 0.5 | 0.01 | 0.001 | 0.0001 | -0.0001 |
---|---|---|---|---|---|---|
slope of secant line | 1.25 | 1.125 | 1.0025 | 1.0025 | 1.00003 | 0.99998 |
From the table, we can guess (but we cannot confirm yet),
\[ 0.99998 \leq \lim_{\Delta x\to 0} \frac{\Delta y}{\Delta x} \leq 1.00003 \]
From the table and the graph, we expect that
\[
1=\lim_{\Delta x \to 0}\frac{\Delta y}{\Delta x}=the~slope~of~tagent~line~at~x_0=2
\]
It means we can make the values of \(\frac{f(x_0+\Delta x)-f(x_0)}{\Delta x}\) arbitrarily to close to a certain number by choosing very small \(\Delta x \neq 0\).
Let \( f(x)=x^2 \). Then \(\lim_{x\to 2} f(x)=?\)
From the graph, we can conclude
\[ \lim_{x\to 2} f(x)=4 \]
Suppose f(x) defined when x is near the number a except \(x=x_0\). (This means that f is defined on some open interval that contains \(x_0\), except possibly at \(x_0\) itself.) Then we write
\[\lim_{x\to x_0}f(x)=L\]
and say "the limit of f(x), as x approaches \(x_0\), equals L".
if we make the values of f(x) arbitrarily close L by taking all x's to be sufficiently close to \(x_0\) but not equal to \(x_0\).
Do not forget, \(x\neq x_0\).
\(f(x_0) = \lim_{x\to x_0}f(x) \)is not true in general.
Actually, \(f(x_0)\) does not need to be defined to define the limit of \(f(x)\) at \(x=x_0\).
1. Continuity | 2. Hole and \(f(x_0)\) is defined. | 3. Hole and \(f(x_0)\) is undefined. |
Let \(f(x)=\frac{x^2-4}{x-2}\). \(\lim_{x\to 2}f(x)=?\)
Let \(x=\frac{1}{n}\) where \(n\) is a natural number. Then
\[f(\frac{1}{n})=\sin(n\pi)=0 \]
So we may guess that \[\lim_{x\to 0} \sin(\frac{\pi}{x})=0\]
But it is wrong. Actually, there is no limit at \(x=0\).