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Consider the function g(x) = cos(1/x). We will investigate the limit behavior at x = 0.c) If n is an arbitrary positive integer, find points x1 and x2 (int terms of n) in the interval(-1/n, 1/n( such that g(x2) = 1 and g(x2) = -1.d) Explain (in a brief paragraph) why (c) implies that g does not have a limit at x = 0.  e) Where is g continuous? Justify your answer. You may use facts from the textbook in Section 2.2 - 2.5.

Question

Consider the function g(x) = cos(1/x). We will investigate the limit behavior at x = 0.

c) If n is an arbitrary positive integer, find points x1 and x2 (int terms of n) in the interval(-1/n, 1/n( such that g(x2) = 1 and g(x2) = -1.

d) Explain (in a brief paragraph) why (c) implies that g does not have a limit at x = 0. 

 

e) Where is g continuous? Justify your answer. You may use facts from the textbook in Section 2.2 - 2.5. 

check_circleAnswer
Step 1

Given function,

 

g(x)
= COS
To investigate the limit behavior at x = 0
If you check the graph of g(x)
2+
-1
the graph of g(x) is very crowded around x=0
help_outline

Image Transcriptionclose

g(x) = COS To investigate the limit behavior at x = 0 If you check the graph of g(x) 2+ -1 the graph of g(x) is very crowded around x=0

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Step 2

The graph of g(x) is very crowded around x = 0.

1 and g(x2) -1
Thus, we find two points, such that g(x
1
So, cos
X1
1
1 then x
_=
2Tn
1
-1 then X1
And cos- =
X2
T(1+2n)
We taken x
2Tt and x2
3Tt
help_outline

Image Transcriptionclose

1 and g(x2) -1 Thus, we find two points, such that g(x 1 So, cos X1 1 1 then x _= 2Tn 1 -1 then X1 And cos- = X2 T(1+2n) We taken x 2Tt and x2 3Tt

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Step 3

d) Now, to check the ...

The function with g(x has limiting value 1 and g(x2) has a limiting
value of -1. So, based on the value of x, and x2, if we define a sequence
1
1
and b
of values an
(2n+1)Ten sequence anwill tend to 1 and
2πη
bnwill tend to -1
Thus, the limit cannot exist for x tending to 0.
help_outline

Image Transcriptionclose

The function with g(x has limiting value 1 and g(x2) has a limiting value of -1. So, based on the value of x, and x2, if we define a sequence 1 1 and b of values an (2n+1)Ten sequence anwill tend to 1 and 2πη bnwill tend to -1 Thus, the limit cannot exist for x tending to 0.

fullscreen

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Tagged in

Math

Calculus

Limits

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