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MATH 161 Quiz 2: Identifying Discontinuities and Finding Limits, Quizzes of Calculus

Millersville university mathematics department's quiz 2 for math 161, focusing on identifying discontinuities and finding limits. Students are required to find the discontinuity of the function f(x) = x^2 - 1 with x - 1, explain their answers, and determine the limit of the function 4 - x / (x - 2)^2 as x approaches 2 from the left.

Typology: Quizzes

Pre 2010

Uploaded on 08/17/2009

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Millersville University Name
Mathematics Department
MATH 161, Quiz 2
January 23, 2004
Please answer the following questions. Your answers will be evaluated on their correctness,
completeness, and use of mathematical concepts we have covered. Please show all work and
write out your work neatly. Answers without supporting work will receive no credit.
1. Find and identify (removeable, jump, or infinite) the discontinuity of the function
f(x) = x21
x1.
Explain your answer fully.
Since f(x) is undefined at x= 1, it must have a discontinuity at x= 1.
We can see that the
lim
x1
x21
x1= lim
x1
(x1)(x+ 1)
x1
= lim
x1(x+ 1)
= 2.
Since the two-sided limit exists as x1, f(x) has a removeable discontinuity at x= 1.
pf2

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Millersville University Name Mathematics Department MATH 161, Quiz 2 January 23, 2004

Please answer the following questions. Your answers will be evaluated on their correctness, completeness, and use of mathematical concepts we have covered. Please show all work and write out your work neatly. Answers without supporting work will receive no credit.

  1. Find and identify (removeable, jump, or infinite) the discontinuity of the function

f (x) =

x^2 − 1 x − 1

Explain your answer fully. Since f (x) is undefined at x = 1, it must have a discontinuity at x = 1. We can see that the

lim x→ 1

x^2 − 1 x − 1

= lim x→ 1

(x − 1)(x + 1) x − 1 = (^) xlim→ 1 (x + 1) = 2.

Since the two-sided limit exists as x → 1, f (x) has a removeable discontinuity at x = 1.

  1. Determine the following limit.

lim x→ 2 −

4 − x (x − 2)^2 As x → 2 −^ the numerator approaches 2. As x → 2 −^ the denominator approaches 0, but is always positive. Thus the fraction is positive as x → 2 −^ and grows to positive infinity. Hence we write lim x→ 2 −

4 − x (x − 2)^2