What is the value of $ \underset{x\to \infty }{\mathop{\lim }}\,\cos x $ ?

Hint: We have to find the limit value of the given function $ \underset{x\to \infty }{\mathop{\lim }}\,\cos x $ . We form the sequence for the function for its domain and get two different values at two different points. Therefore, the limit of the given function doesn’t exist.

Complete step by step solution:
The given limit value $ \underset{x\to \infty }{\mathop{\lim }}\,\cos x $ doesn’t exist.
The function $ f\left( x \right)=\cos x $ is a continuous function. As the fixed value for $ x $ is not mentioned, the range of the function \[-1\le \cos x\le 1\]. The limit value will be in the whole range with no limit value.
For the given function $ \underset{x\to \infty }{\mathop{\lim }}\,\cos x $ , we take a sequence in the form of $ {{x}_{N}}=2\pi N,N\to \infty $ . We get that for any value of $ N $ , we have $ f\left( x \right)=\cos \left( {{x}_{N}} \right)=1 $ .
We can also take a sequence in the form of $ {{x}_{N}}=\dfrac{\pi }{2}+2\pi N,N\to \infty $ . We get that for any value of $ N $ , we have $ f\left( x \right)=\cos \left( {{x}_{N}} \right)=0 $
So, the first sequence of values of $ f\left( x \right)=\cos \left( {{x}_{N}} \right) $ equals 1 and the limit must be 1. The second sequence equals to 0 and limit must be 0. But the limit cannot be simultaneously equal to two distinct numbers. Therefore, there is no limit.

Note: The inversion form of the limit also can prove the same thing. We can write in the form of $ \underset{x\to \infty }{\mathop{\lim }}\,\cos x=\underset{z\to 0}{\mathop{\lim }}\,\cos \left( \dfrac{1}{z} \right) $ by the inversion of $ z=\dfrac{1}{x} $ .