Is it true that lim x->(infinity) cos x = lim x->(-infinity) cos x?

Question

Is it true that lim x->(infinity) cos x = lim x->(-infinity) cos x?

Look at the graph of f(x) = cos x.

^_^

Answer 1

It's true that cos(x) = cos(-x_ for every real x. But the limits you geve do'nt exist. The cosine function is periodic and it's values oscillates from -1 to 1. Whem x -> oo or -oo, cos (x) doesn't approach any real value.

Answer 2

Well the cosine graph is an even function, which means that it is reflected in the y-axis. So technically as you move x towards the limits the cosine function will indeed remain the same.

However, the cosine function has no definate value at +-infinity as it oscillates forever in both directions. But if you approached the limits in exactly the same way then the functions would remain identical

Answer 3

OK for the fact that for every x, cos(x) = cos(-x). As long as x goes towards the infinite, the value of cos(x) is defined and it's possible to consider it. When x becomes infinite, the cos(x) value is undefined. There is no such thing as lim (for x -> infinity) cos(x).
Your relation is not true for that reason.

Answer 4

yes, but technically.

for clear understanding, you need to realise the fact about limit.
lt x->α refers to the fact that x is having a very large value but a real one. that means for (lt. x->α) x surely possesses a real value.

now whatever the very large value is, plotting on quardents u find them either as (2nπ+-x) or, {(2n-1)π+-x}. the first angle resides in 1/4 quardent and the second angle resides in 2/3 quardents. as you know (cos x) is +ve in 1/4 quardent and -ve in the 2/3 quardent and as known for a certain value if x is in 1/2 quardent, for the same negative value x is in the 4/3 quardent or vice versa, lim x->(infinity) cos x = lim x->(-infinity) cos x is definitely true.

also its easy to understand from stuart_T's logic. so why to hesitate?

Answer 5

Graph f(x) = cos -x and if the graphs are the same then it is true