Prove that $\lim\limits_{n\to \infty} \cos(a+b/n)=\cos(a)$ by the definition of limits, where $a$ and $b$ are positive numbers My approach:
For all $\epsilon \gt 0$, we need to find $K$ in natural numbers, s.t.
$$|\cos(a+b/n)-\cos(a)| <\epsilon, \text{ for all } n\ge K.$$
I want to convert $|\cos(a+b/n)-\cos(a)| <\epsilon$ to a form that $n$ is in the L.H.S. and $\epsilon$ is in the R.H.S. in order to find the $K$.
I don’t know what to do for this step.

Alternative approach
First of all, you can use the fact that 
$\displaystyle 
\lim_{x\to 0} \frac{\sin(x)}{x} = 1$ 
to conclude that 
$\displaystyle \lim_{x \to 0} \sin(x) = 0.$
Further, you can use the Taylor series of 
$\displaystyle \cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots - \cdots$ 
to conclude that 
$\displaystyle \lim_{x \to 0} \cos(x) = 1.$
Alternatively, these same preliminary results are established if you are allowed to assume that the sine and cosine functions are each continuous at $x = 0$.

Then:
$$\cos\left(a + \frac{b}{n}\right) = \cos(a)\cos\left(\frac{b}{n}\right) - \sin(a)\left(\frac{b}{n}\right). \tag1 $$
As $\displaystyle n \to \infty, ~\dfrac{b}{n} \to 0 \implies \cos\left(\frac{b}{n}\right) \to 1.$
Similarly, as $\displaystyle n \to \infty, ~\dfrac{b}{n} \to 0 \implies \sin\left(\frac{b}{n}\right) \to 0.$
Then, it is clear that as $n \to \infty$, the RHS of (1) above goes to
$$[\cos(a) \times 1] - [\sin(a) \times 0].$$
