# Prove that lim sin(x)/x = 1 as x goes to 0(Epsilon delta )

#### Amer

##### Active member
Prove that

$$\lim_{x\rightarrow} \frac{\sin x}{x} = 1$$

Solution
Given $$\epsilon > 0$$
want to find $$\delta$$ such that $$\left|\frac{\sin x}{x} - 1 \right| < \epsilon$$
for x, $$|x | < \delta$$

can I use Taylor expansion of sinx ? but Taylor is an approximation of sin(x) around a certain point ? how to find such a delta ?
Thanks

#### Siron

##### Active member
$$\left|\frac{\sin(x)}{x}-1\right| = \left|\frac{\sin(x)-x}{x}\right| = \frac{|\sin(x)-x|}{|x|} \leq \frac{|\sin(x)|+|x|}{|x|}$$

Note that $|\sin(x)|\leq 1$ therefore we can use the following upper bound
$$\frac{|\sin(x)|+|x|}{|x|}\leq \frac{1+|x|}{|x|}<\epsilon$$

At this point we can't to get rid of the $|x|$ in the denominator ($\epsilon$ can not be depending on $x$). We need another upper bound for $|x|$, note that $x$ has to lie in the neighbourhoud of $0$ thus it's allowed to say that $|x|<1$.

Can you make a conclusion now?

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