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# lim(x/(2-cos(x)),x->infinity

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lim(x/(2-cos(x)),x->infinity

Dec 26, 2015

#5
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The denominator varies between 1 and 3  (finite positive numbers).  The numerator tends to infinity.  Infinity divided by any finite number is still infinite.

Dec 27, 2015

#1
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Find the following limit:
lim_(x->infinity) x/(2-cos(x))

Applying the quotient rule, write lim_(x->infinity) x/(2-cos(x)) as (lim_(x->infinity) x)/(lim_(x->infinity) (2-cos(x))):
(lim_(x->infinity) x)/(lim_(x->infinity) (2-cos(x)))

The limit of a difference is the difference of the limits:
(lim_(x->infinity) x)/lim_(x->infinity) 2-lim_(x->infinity) cos(x)

Using the fact that cosine is a continuous function, write lim_(x->infinity) cos(x) as cos(lim_(x->infinity) x):
(lim_(x->infinity) x)/(lim_(x->infinity) 2-cos(lim_(x->infinity) x))

lim_(x->infinity) x  =  infinity:
(lim_(x->infinity) x)/(lim_(x->infinity) 2-cos(infinity))

cos(infinity) = -1 to 1:
(lim_(x->infinity) x)/lim_(x->infinity) 2--1 to 1

Since 2 is constant, lim_(x->infinity) 2  =  2:
(lim_(x->infinity) x)/(2--1 to 1)

2--1 to 1 = 1 to 3:
(lim_(x->infinity) x)/1 to 3

lim_(x->infinity) x  =  infinity:
infinity/(1 to 3)

infinity/(1 to 3) = infinity:

Dec 26, 2015
#2
+8

Look att he graph of the function, here.......https://www.desmos.com/calculator/m8gxiqxa9c

Would it be more accurate to say that the limit does not exist.......since the function "ping-pongs"  between values defined by two non-parallel lines as x approaches infinity    ???   Dec 26, 2015
#3
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CPhill: Just plugged it into WolframAlpha and it also gives infinity????!!!!!!!!.

Dec 26, 2015
#4
+5

OK......thanks, guest.....I stand corrected......!!!!   Dec 26, 2015
edited by CPhill  Dec 27, 2015
#5
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The denominator varies between 1 and 3  (finite positive numbers).  The numerator tends to infinity.  Infinity divided by any finite number is still infinite.

Alan Dec 27, 2015
#6
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Thanks, Alan......!!!!!   Dec 27, 2015