Limit: $\displaystyle{\lim_{x \to c}}$ $f(x) = L$
$c = x$ value and $L = y$ value
A limit is arbitaritly number close to $L$ by testing numbers around $c$ on it’s left and right sides.
Ex 1: Evaluate the function $f(x)=\frac{(x^2-4)}{(x-2)}$ at values near $x=2$ to estimate $\displaystyle{\lim_{x \to 2}}$ $f(x)$.
$\displaystyle{\lim_{x \to 2}} f(x)={\frac{(x^2-4)}{(x-2)}} = \frac{(2)^2-4}{2-2} = \frac{0}{0}$
$x$ | $f(x)$ |
---|---|
1.9 | 3.9 |
1.99 | 3.99 |
1.999 | 3.999 |
2.001 | 4.001 |
2.01 | 4.01 |
2.1 | 4.1 |
So the limit of $x$ approaching 2 is about 4.
Find the limit as $x$ approachs to 3:
$\displaystyle{\lim_{x \to 3}} f(x)={\frac{(x^2-4)}{(x-2)}} = \frac{(3)^2-4}{3-2} = \frac{9-4}{3-2} = 5$
Ex 2: Estimate the limit $\displaystyle{\lim_{x \to 0}}$ where
$f(x)=\begin{cases}\frac{\cos x -1}{x} & if x\neq0 \ 4 & if x = 0 \ \end{cases}$
$f(0) = 4$
It’s possible for the limit not to match the actual point of the function:
$\displaystyle{\lim_{x \to c}} f(x)\neq f(c)$
$x$ | $f(x)$ |
---|---|
-0.5 | 0.2448 |
-0.1 | 0.04996 |
-0.01 | 0.0049 |
0.01 | -0.0049 |
0.1 | -0.04996 |
0.5 | -0.2488 |
So the limit of $x$ approaching 0 is about 0.
$\displaystyle{\lim_{x \to 0}} f(x) = 0 $
Ex 3: Estimate $\displaystyle{\lim_{x \to 1}}$ $f(x)$ $\frac{1}{x^2-2x+1}$
$x$ | $f(x)$ |
---|---|
0.9 | 100 |
0.99 | 10000 |
0.999 | 1000000 |
1.001 | 1000000 |
1.01 | 10000 |
1.1 | 100 |
So the limit of $x$ approaching 1 does not exist. You can see this graphically:
Ex 4: Find $\displaystyle{\lim_{x \to 0}}$ $\frac{\lvert x\rvert}{x}$ using its graph then by simplifying the equation.
This limit does not exist since the left and the right have different values.
Ex 5: Fill in the tables of values to estimate the limit: $\displaystyle{\lim_{x \to 0}}$ $\sin \frac{1}{x}$
$x$ | $\sin \frac{1}{x}$ | $x$ | $\sin \frac{1}{x}$ | $x$ | $\sin \frac{1}{x}$ |
---|---|---|---|---|---|
2/$\pi$ | 1 | 2/3$\pi$ | -1 | 1/$\pi$ | 0 |
2/5$\pi$ | 1 | 2/7$\pi$ | -1 | 1/5$\pi$ | 0 |
2/13$\pi$ | 1 | 2/31$\pi$ | -1 | 1/10$\pi$ | 0 |
2/101$\pi$ | 1 | 2/99$\pi$ | -1 | 1/100$\pi$ | 0 |
The limit does not exist because the function continually oscillates between -1 and 1 as $x\ge0$.