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Calculus Task 1

The document discusses limits and continuity for the function f(x) = (x−2)/(x−3)(x+2). It finds that the limit does not exist as x approaches 1 due to an undefined value, while the limit as x approaches 2 is 1. It identifies points of discontinuity at x=1 and x=2, affecting the domain and range of the function by excluding these values.

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jeffjmartin
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72 views2 pages

Calculus Task 1

The document discusses limits and continuity for the function f(x) = (x−2)/(x−3)(x+2). It finds that the limit does not exist as x approaches 1 due to an undefined value, while the limit as x approaches 2 is 1. It identifies points of discontinuity at x=1 and x=2, affecting the domain and range of the function by excluding these values.

Uploaded by

jeffjmartin
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as DOCX, PDF, TXT or read online on Scribd
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Jeffrey Martin

Calculus I – QJT2
QJT Task 1: Limits and Continuity

x−2 x −2
A. Given f(x) = 2 , find lim 2 or explain why the limit does not exist.
x −3 x +2 x →1 x −3 x +2

−1
By plugging in 1 to the equation we get which is undefined therefore the limit does not exist
0
(see work below):
x −2 1−2 −1
lim 2 = 2 = and when we factor we get:
x →1 x −3 x +2 1 −3(1)+2 0
x −2 x−2 1 1 1
lim 2 = lim = lim = = which is still undefined therefore the
x →1 x −3 x +2 x →1 ( x−1)(x−2) x →1 x−1 1−1 0
limit does not exist.

x−2 x −2
B. Given f(x) = 2 , find lim 2 or explain why the limit does not exist.
x −3 x +2 x →2 x −3 x +2

The limit of x as it approaches 2 is 1. As we can see in the work below, by plugging in 2 we get
0
so we need to try factoring, and when we factor we get 1.
0

x −2 2−2 0 0
lim 2 = 2 = =
x →2 x −3 x +2 2 −3(2)+2 4−4 0

x −2 x−2 1 1 1
lim 2 = lim = lim = = =1
x →2 x −3 x +2 x →2 ( x−1)(x−2) x →2 ( x−1) 2−1 1

C. Apply the definition of continuity to identify any points of discontinuity in the function f(x)
above, showing all work.
For a function to be continuous it must be defined at that point, the limit must exist at that point
and the function value and the limit value must be equal.
x−2 x−2
f(x) = 2 when we factor the equation we get lim and as you can see from
x −3 x +2 x →2 ( x−1)(x−2)

the work above, if x= 1 or 2 that will give us a zero in the denominator which means the function
is undefined so when x approaches 1 the first part of definition fails. Also when x approaches 2
we get a zero in the denominator which is also undefined so when x approaches 2 the first part of
the definition fails.
1. Explain the effect of the discontinuities identified in part C on the domain and range of the
function f(x) above.
The domain will be changed from all real numbers to all real number except x=1 and x=2. You
can see this if you graph the function when the range is y=0 there is a horizontal asymptote so
this excludes 0 from the range.
As you can see from the work above, when x=2 the limit is 1. Since this creates a hole at x=2,
y=1 must also be excluded from the range.
So the range will be all real numbers except y=0 and y=1.

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