Limits
In order to really understand Calculus, we first must understand a limit of a function at a point. The notion of a limit is the value the function approaches (y) as we approach a particular value of x. We will look at it from a graph, a table and algebraically with a piecewise function.
[expand title="Limits from a Graph" trigclass=""]
A.
$latex begin1.,,text,,text,,text,,text,,underset,F(x),=,2.,,text,,text,,text,,text,,text,,text,,text,,text,,text,,text,,\text,,text,,text,,text,,text,,text,,text,,text,,text,,text,,text,,text,,text,,text,,text\text,,text,,text,,text,,text,,text,,text,,text,,text,,text,,text,,text,,text,,text,,textend$
$latex begin2.,,text,,text,,text,,text,,underset,,G(x),=1,,text,,underset,,G(x),=3.,,text,,text,,text,,text,,text,,text,,text,,text,text,,text,,text,,text\text,,text,,underset,,G(x),ne underset,,G(x),,,,text,,text,,text,,text,text,\text,,underset,,,G(x),,text,,text,,textend$
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[expand title="Limits from a Table" trigclass=""]
Similarly, when we look at function values from a table, we need to look at the value the function APPROACHES from both sides, not the value at the point.
A. $latex Find,,underset,,,F(x)$
$latex begintext,,,underset,F(x)=4,,,text,,underset,F(x)=4,,,text,,underset,F(x)=4.,,text,,text,,text,,text,,text\F(x),,text,,text,,text,,text,,x=3,,text,,text,,text,,text,,text,,text,,text,,text,,textend$
B. $latex Find,,underset,,,G(x)$
$latex text,,,underset,G(x)=4,,,text,,underset,G(x)=-4,,,text,,underset,G(x),,text,,text,,text$
C. $latex Find,,underset,,,K(t)$
$latex begintext,,,underset,K(t)=3,,,text,,underset,K(t)=3,,,text,,underset,K(t)=3.,,text,,text,,text,,text,,text\K(-4)=0,text,,text,,text,,text,,text,,text,,textend$
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[expand title="Limits Algebraically for a Piecewise Function" trigclass="blue"]
A. $latex begintext,,underset,,,text\text=left{ beginx-3,,,,,,text,,x<4\2,,,,,,,,,,,,,,text,,x=4\2x-7,,,,text,,x>4end right.\\text,,,\underset,F(x)=underset,x-3=1\underset,F(x)=underset,2x-7=1\\text,,text,,text,,text,,text,,text,,text,,text,,text,,text,,text=text,,text,,text,,text,,text,,text\text,,text,,text,,text,,text,,text,,text,,text,,text,,text,,\text,,text,,text,,text,,text,,text,,text,,text,,textend$
B. $latex begintext,,underset,,,text\text=left{ begin{^{2}}+4,,,,,,when,,text
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Continuity
The next important fundamental concept is continuity or determining if a function is continuous. In order for a function to be continuous at a particular point, it must first have a limit at the point – if it does not have a limit, it is not continuous. Secondly, the value of the function must also equal the limit. More formally, $latex underset,F(x)=underset,F(x)=F(A)$ In other words, there cannot be any breaks, jumps or gaps in the function or graph. More informally, a simple way to remember continuity is that when drawing the graph, you never lift your pencil when a function is continuous. If you lift your pencil, it is not a continuous function. Please note that a function can be continuous for all values on an interval, except for a single point.
[expand title="Continuity from a Graph" trigclass=""]
A.
$latex begintext,,text,,text,,text,,,text,,,underset,,F(x)=,2.,,text,,text,,text,,text,,text,,text,,text,,text=text,,text,,text,,text,,text,,text\text=text,,text,,text,,text,,text,,text,,text,,text,,text,,text,,text,,mathbf,,text,,text,,text=text\text,,text,,text,,,text,,,underset{text,to ,text},,F(x)=,2.,,Ttext,,text,,text,,text,,text,,text,,text=text,,text,,text,,text,,2,,text\text=2text,,text,,text,,text,,text,,text,,text,,text,,text,,text,,,text,,text,,text=textend$
2. Using the picture to the left, no limit exists at A, so it is not continuous at x = A. The function is continuous on the interval [Q , A) and [A , R], but not at x = A
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[expand title="Continuity from a Table" trigclass=""] Similarly, when we look at function values from a table, we need to check that it has a limit and then verify that the limit equals the value of the function.
A. Is F(x) continuous at x = 3?
Using the picture above, the limit as x approaches 3 is 4, written $latex ,underset,,F(x)=,4$ . However, the limit does not equal the function value at x = 3, since F(x) is undefined at 3. Hence, F(x) is NOT continuous at x = 3.
B. Is G(x) continuous at x = 3?
Using the picture above, no limit exists at x = 3, since the Right-Hand Limit and the Left-Hand Limit are NOT equal, written: $latex ,,underset,,G(x),,=,,4,,,,and,,,,underset,,G(x),,=,,-4,,,so,,underset,,G(x)textne underset,,G(x)$ Since no limit exists x = 3, G(x) is NOT continuous at x = 3.
C. Is K(t) continuous at t = -4?
Using the picture above, the limit as t approaches t = -4 is 3, $latex ,,underset,,K(t)=,underset,,K(t)=,4$ However, the limit does not equal the function value at t = 3, so K(t) is NOT continuous at t = 3.
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[expand title="Continuity Algebraically for a Piecewise Function" trigclass="blue"]
A. $latex begintext,,text,,text,,text,,text=text\text=left{ begintext-text,,,,,,text,,text
B. $latex begintext,,text,,text,,text,,text=6text\text=left{ begin{{text}^{text}}-text,,,,,,text,,text
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Differentiability
The final fundamental concept is differentiability or determining if a function is differentiable. In order for a function to be differentiable at a particular point, it must be continuous (hence a limit also) – if it is not continuous, it is not differentiable. Secondly, the slope of the function must be the same as we approach from both sides. More formally, $latex underset,text=underset,text,,text,,text,,text,,text,,text,,text=text.$ In other words, only 1 tangent line can be draw at a given point if a function is differentiable. More informally, a simple way to remember differentiability is that the graph is smooth and has no sharp points. Special Notes:
1. If a vertical asymptote exists at a point, it is NOT continuous and therefore not differentiable either.
2. The absolute value function is continuous at all points and differentiable at all points except at the bottom (or top if it is negative) of the v-shaped graph.
[expand title="Differentiability from a Graph" trigclass=""]
A. Where is the function F(x), shown below, differentiable?
The function F(x) to the left is NOT differentiable at A, because it is NOT continuous at A. The function F(x) is differentiable at B, because the slope approaches the same value from each side or you can only draw one tangent line to the graph at x = B. In addition, F(x) is not differentiable at either the left or the right endpoint. The slope must approach the same value from the left and right. The slope at the left endpoint only has a right sided limit and the slope at the right endpoint only has a left sided limit so a function is NEVER differentiable at the end points.
2. Using the picture to the left, no limit exists at A, so it is not continuous at x = A. The function is continuous on the interval [Q , A) and [A , R], but not at x = A
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[expand title="Determining the Differentiability for a Piecewise Function, Algebraically" trigclass="blue"]
A. $latex begintext,,text,,text,,text,,text=text\text=left{ begintext-text,,,,,,text,,text B. $latex begintext,,text,,text,,text,,text=6text\text=left{ begin{{text}^{text}}-text,,,,,,text,,text [/expand]
