I'm not out to trick you

$-1\leq \cos x \leq 1$

$\displaystyle \Rightarrow 0 \leq 1-\cos x \leq 2$

Since the numerator is finite for any real $x$ we can say that:

$\displaystyle \lim_{x\rightarrow +\infty} \frac{1-\cos x}{x} = k \times \displaystyle \lim_{x \rightarrow + \infty} \frac{1}{x} ; k \in [0,2]$

$\displaystyle \Rightarrow \displaystyle \lim_{x\rightarrow +\infty} \frac{1-\cos x}{x} = k \times 0 = \boxed{0}$

We rewrite $\large 1 - \cos x$ as $\large 2 \sin^2 \frac{x}{2}$ .

Observe that $\large 2 \sin^2 \frac{x}{2}$ can attain a maximum value of 2 and a minimum of zero. As x approaches infinity, this it keeps on oscillating between its lower and upper bounds.

Thus, the limit would tend to zero as x goes to infinity.

Note that $0\leq\left|\frac{1-\cos x}{x}\right|\leq\frac{1+|\cos x|}{|x|}\leq\frac{2}{|x|}$

Now, taking $x\rightarrow+\infty$ , we conclude by the squeeze theorem/sandwich theorem that $\lim_{x\rightarrow+\infty}\frac{1-\cos x}{x}=\boxed{0}$