The limit does not exist because as x approaches -4 from the left the limit is 0 but as x approaches -4 from the right the limit is 2. You cannot have two answers, therefore the limit does not exist.
14
Expression 15: left brace, "x" less than negative 4 : sin left parenthesis, "x" plus 4 , right parenthesis , "x" greater than or equal to negative 4 : negative 1 half "x" , right bracex<−4:sinx+4,x≥−4:−12x
15
Does the following function have a limit at a=0? If so, what is it?
16
The limit of f(x) as x approaches 0 is undefined but with estimations the limit is 1.
17
Expression 18: StartFraction, sin left parenthesis, "x" , right parenthesis Over "x" , EndFractionsinxx
18
Does the following function have a limit at a=0? If so, what is it?
19
The limit of f(x) as x approaches 0 is undefined but with estimations the limit can be approximated to be 1.
20
Expression 21: sine left parenthesis, StartFraction, 1 Over "x" , EndFraction , right parenthesissin1x
21
The next function has a constant k in it that you can change with the slider. Find a value for k so that the limit of the function exists at every point. Answer: k=
22
k=2
23
Expression 24: "f" left parenthesis, "x" , right parenthesis equals StartFraction, "x" squared minus "x" minus 2 Over "x" minus "k" , EndFractionfx=x2−x−2x−k
24
Expression 25: "k" equals 3k=3
negative 10−10
1010
25
Why does this value for k work? Explain what you see geometrically, and if possible explain algebraically. Answer:
26
The value of k=2 works because 2 is an intercept. When approaching from both the left ad the right the limit always goes to 0.
27
When you are done, save this graph and share it with me (dmarshall@nmhschool.org).
