The real limit of a function f(x)f(x), if it exists, as x->oox→∞ is reached no matter how xx increases to oo∞. For instance, no matter how xx is increasing, the function f(x)=1/xf(x)=1x tends to zero.
This is not the case with f(x)=cos(x)f(x)=cos(x).
Let xx increases to oo∞ in one way: x_N=2piNxN=2πN and integer NN increases to oo∞. For any x_NxN in this sequence cos(x_N)=1cos(xN)=1.
Let xx increases to oo∞ in another way: x_N=pi/2+2piNxN=π2+2πN and integer NN increases to oo∞. For any x_NxN in this sequence cos(x_N)=0cos(xN)=0.
So, the first sequence of values of cos(x_N)cos(xN) equals to 11 and the limit must be 11. But the second sequence of values of cos(x_N)cos(xN) equals to 00, so the limit must be 00.
But the limit cannot be simultaneously equal to two distinct numbers. Therefore, there is no limit.
