#1**+10 **

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:

**Answer: | infinity**

Guest 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 ???

Anyone else have any thoughts about this question ????

CPhill
Dec 26, 2015

#3**+10 **

CPhill: Just plugged it into WolframAlpha and it also gives infinity????!!!!!!!!.

Guest Dec 26, 2015