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How do I prove that $\lim_{x \to 0} \sin(x) = 0$ using the episilon-delta definition of a limit?

Do I have to divide the domain of $x$ into 4 cases for each quadrant?

Update:

Based on the input from @vadim123,

For any $\epsilon>0$ if there exists some $\delta>0$ satisfying $|x-0|<\delta$, then we have to show that $|\sin\ x-0|<\epsilon$

Choose $\delta=\epsilon$.

By transitivity of the $<$ and $\le$ relations, $|\sin\ x |\le|x|<\epsilon$.

we have now shown that $|\sin\ x| < \epsilon$.

Martin Sleziak
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takwing
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    The tools one is allowed to use are not clear to me, it depends on how $\sin x$ is defined. Informally, with the "geometric" definition, we have $|\sin x|\le |x|$, and then there is little to do. – André Nicolas Nov 13 '13 at 15:49
  • The tag ([tag:limit-theorems]) is intended for questions about limit theorems in probability theory and not for questions about determining limits of sequences or functions, see the tag-wiki and the tag-excerpt. (The tag-excerpt is also shown when you are adding a tag to a question.) – Martin Sleziak Dec 02 '13 at 09:33

2 Answers2

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Hint: $|\sin x|<|x|$, for all $x$.

vadim123
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For fun, let's make a geometric proof.

Let $O$ be the origin. Draw a unit circle $C$ centered at the origin. Recall that $\sin \theta$ can be defined as the $y$ coordinate of the point on the unit circle lying $\theta$ units counter-clockwise from the positive $x$-axis.

Suppose $\epsilon > 0$. Our goal is to find $\delta > 0$ such that every $\theta$ such that $0 < |\theta| < \delta$ implies $|\sin \theta| < \epsilon$.

Consider the horizontal lines $\ell$ and $\ell'$ defined by $y = \epsilon$ and $y = -\epsilon$. Every point that is on the unit circle whose $y$ coordinate is between $-\epsilon$ and $\epsilon$ lies between these two lines.

If $\epsilon < 1$, $\ell$ intersects the circle in two points.

We may assume $\epsilon < 1$. (do you understand why?)

Of these two points, let $Q$ be the point with positive $x$ coordinate. Similarly, let $Q'$ be the point where $C$ meets $\ell'$ with positive $x$ coordinate.

Note that every point on the unit circle that has positive $x$ coordinate and has $y$ coordinate satisfying $|y| < \epsilon$ lies within the angle $\angle QOQ'$.

Let $\delta$ be the measure of $\angle QOP$. All of the above shows that if $|\theta| < \delta$, we have $|\sin \theta| < \epsilon$.