# Could you please prove that lim_(x rarr 0) sin(x)/x = 1 using the formal definition of limits?

Apr 15, 2017

$\forall \epsilon > 0 , \exists \delta = \sqrt{6 \epsilon} > 0$ s.t. 0 < |x| < delta Rightarrow|sin x/x-1| < epsilon

#### Explanation:

Recall:

1. Error Bound for Alternating Series
Let $s = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {b}_{n}$, where ${b}_{n} > 0$ and ${b}_{n}$ is monotonically decreasing, and let
${s}_{n} = {\sum}_{i = 0}^{n} {\left(- 1\right)}^{n} {b}_{n}$.
The error for approximating the sum $s$ using the partial sum ${s}_{n}$ is bounded by ${b}_{n + 1}$, that is, $| s - {s}_{n} | \le q {b}_{n + 1}$

2. Power Series of sin x
sin x =x-x^3/(3!)+x^5/(5!)-x^7/(7!)+cdots

So, the error bound for estimating $\sin x$ by the first term $x$ can be expressed as:
|sin x -x| leq |x^3/(3!)|

Let us now prove the limit.

$\forall \epsilon > 0 , \exists \delta = \sqrt{6 \epsilon} > 0$ s.t. $0 < | x - 0 | < \delta R i g h t a r r o w | x | < \sqrt{6 \epsilon}$

By Error Bound for Alternating Seires,

Rightarrow |sin x/x -1|=|(sin x - x)/x|leq|x^3/(3!)|/|x|=x^2/6<(sqrt(6epsilon))^2/6=epsilon

Hence, ${\lim}_{x \to 0} \sin \frac{x}{x} = 1$