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