Exploring Limits with Graph Continuity at a Point Math 1241, Spring 2014 Section 3.2 Exploring Limits with Graph Continuity at a Point

Exploring Limits with Graph We’ll start with a simple example: using Graph to study lim 𝑥→2 2𝑥−3 =1 . Start by graphing 𝑓 𝑥 =2𝑥−3. Add the point (2,1) to the graph. Zoom in on the point (2,1), using either: With the cursor close to (2,1), scroll up on the mouse wheel. Using the rectangle zoom, drag a small rectangle around (2,1). No matter how far you zoom in, the limit seems correct. To reset the view, use <CTRL>-<D> and then <CTRL>-<Q> to return to a standard view centered at the origin.

Tracing Functions in Graph Highlight the function 𝑓 𝑥 =2𝑥−3 in the left pane of the Graph window. Select Calc -> Evaluate from the top menu (or use the button on the toolbar). Make sure “Snap to” is set to “Function.” Click/drag the mouse near the limit point to see the nearest point on the graph. Note the values of x and f(x) the lower left.

Tables of Values in Graph Highlight the function 𝑓 𝑥 =2𝑥−3 in the left pane of the Graph window. Select Calc -> Table from the top menu (or use the button on the toolbar). In the new window, set the following values: From = 1, To = 3, Δx = 0.1 Click the “Calc” button to generate the table.

Tables of Values in Graph Try the following values (click the “Calc” button each time to regenerate the table) From = 1.9, To = 2.1, Δx = 0.01 From = 1.99, To = 2.01, Δx = 0.001 From = 1.999, To = 2.001, Δx = 0.0001 Note how values of f(x) are close to the limit value (1) as x is close to the target value (2).

Exercises Use Graph to explore the following: lim 𝑥→2 𝑥 3 −3 𝑥 2 +2𝑥+1 =1 lim 𝑥→0 𝑥 3 −3 𝑥 2 +2𝑥+1 =1 lim 𝑥→−1 𝑥 3 −3 𝑥 2 +2𝑥+1 =−5 Note: Uncheck the boxes in the left-hand pane if you want to temporarily hide previous graphs.

Exercise: Rational Functions Let 𝑓 𝑥 = 𝑥 2 −2𝑥 𝑥 2 +𝑥−6 . Use Graph to study: lim 𝑥→1 𝑓(𝑥) = 1 4 lim 𝑥→2 𝑓 𝑥 = 2 5 Note: When typing the function into Graph, be sure to use appropriate parentheses—the entire numerator and entire denominator should each be within a set of parentheses.

𝑓 𝑥 = 𝑥 2 −2𝑥 𝑥 2 +𝑥−6

Removing Zero Denominators Last time, we saw that: 𝑓 𝑥 = 𝑥 2 −2𝑥 𝑥 2 +𝑥−6 = 𝑥(𝑥−2) (𝑥−2)(𝑥+3) Although the graph does not show that f(2) is undefined, the table of values reveals this! Be sure to graph the simplified version 𝑥 𝑥+3 . You will see that the two graphs overlap.

Infinite Limits Let 𝑓 𝑥 = 𝑥 2 −2𝑥 𝑥 2 +𝑥−6 . Use Graph to study: You can see the behavior on the graph. A table of values can help to confirm what you see. Try From = - 3.1, To = - 2.9, and Δx = 0.01.

Continuity (Section 3.2) Very often, we evaluate lim 𝑥→𝑎 𝑓(𝑥) by plugging in x = a and computing the value of f(a). If lim 𝑥→𝑎 𝑓 𝑥 =𝑓(𝑎) , we say that the function f(x) is continuous at the point x = a. At or near the value x = a, there are no breaks or jumps in the graph. We could trace the graph without picking up our pen/pencil.

Discontinuities If lim 𝑥→𝑎 𝑓 𝑥 ≠𝑓(𝑎) , we say that the function f(x) has a discontinuity at the point x = a. There are several possible causes of this: The function value f(a) is undefined. lim 𝑥→𝑎 𝑓 𝑥 does not exist (possibly because the limit “equals” +∞ or −∞). The function value does not equal the limit value (this is rare for “ordinary” algebraic functions).

Example: Discontinuities The function shown has discontinuities at: x = 6 (function value is undefined) x = -4 (limit does not exist) x = 1 (function value is not equal to the limit)

One-sided Continuity When lim 𝑥→𝑎 𝑓(𝑥) does not exist, it might still be the case that f(a) is equal to either the left- or right sided limit. Previous slide: lim 𝑥→− 4 − 𝑓 𝑥 =3=𝑓 −4 . If f(a) is equal to the left-sided (or right-sided) limit at x = a, we say that f(x) continuous from the left (or from the right) at the point x = a.

Exercise: One-sided Continuity Where is the function shown discontinuous? At each discontinuity, is the function continuous from either the left or from the right? Pay attention to the open and closed circles on the graph!