Finding a limit using a definition I want to prove, for $a,b$ positive numbers that $$\lim_{n\to \infty} \cos \left( a + \frac{b}{n}\right) = \cos a$$ by using the definition of a limit. In order to prove this, can I first prove by letting $a_n$ be the sequence which equals $1/n$, where $n = 1,2,3,\ldots$ then prove $a_n \to 0$ as $n$ approaches infinity, then apply this sequence $a_n$ to this equation like $$\lim_{n\to\infty} \cos \left( a + b\cdot a_n \right) = \cos \left( a + b \cdot \lim_{n\to\infty} a_n \right) = \cos(a)$$ as $n$ approaches infinity?
 A: A similar approach is to use Taylor expansions of sine and consine funtions:
$\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$,    $\cos(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots$ to prove
$\lim_{x\to 0}\cos x=1$, $\lim_{x\to 0}sin x=0$
Then,
$\lim_{n\to \infty} \cos \left( a + \frac{b}{n}\right) $
=$ \lim_{n\to \infty} (\cos  a\cos\frac{b}{n}-\sin  a\sin\frac{b}{n})$
=$ \lim_{x\to 0} (\cos  a\cos x-\sin  a\sin x)$
=$ \cos  a  (\lim_{x\to 0}\cos x)-\sin  a( \lim_{x\to 0}sin x)$ (cosine function is continuous)
=$ \cos  a.1-\sin  a.0$
=$\cos a$
A: That's correct but you should mention why $a_n\to 0\implies \cos (a+ba_n)\to\cos a$. That is because of continuity of $\cos$ function. But let's pretend that we don't know about continuity of $\cos$ function. If $b=0$, we are done so assume $b\ne 0$.
$a_n\to 0\implies \forall \epsilon\gt 0, \exists N\in \mathbb N: n\gt N\implies|a_n|\lt \epsilon/|b|$ 
For $n\gt N$, we also have 
$|\cos (a+ba_n)-\cos a|=2|\sin \frac{a+ba_n}{2}\sin\frac{ba_n}{2}|\le|2 (1)\sin \frac{ba_n}{2}|\le |ba_n|\lt \epsilon$. 
This proves the result by definition.
A: $$\lim_{n\to \infty} \cos \left( a + \frac{b}{n}\right) $$
$$\cos(a+\frac{b}{n} ) = \cos(a)\cos(\frac{b}{n})-\sin(a)\sin(\frac{b}{n})$$
$$\lim_{n\to \infty} \big(\cos(a)\cos(\frac{b}{n})-\sin(a)\sin(\frac{b}{n})\big)$$
$$\lim_{n\to \infty} (\dots) = \lim_{\frac{b}{n} \to 0} (\dots)$$
$$\lim_{\frac{b}{n} \to 0} \big(\cos(a)\cos(\frac{b}{n})-\sin(a)\sin(\frac{b}{n})\big)$$
$$\lim_{x\to 0} \sin(x) = 0$$
$$\lim_{x\to 0} \cos(x) = 1$$
$$\lim_{\frac{b}{n} \to 0} \big(\cos(a)\cos(\frac{b}{n})-\sin(a)\sin(\frac{b}{n})\big)$$
$$\lim_{\frac{b}{n} \to 0} (\cos(a)-0)$$
$$\cos(a)$$
