# Question about the proof of a limit

Here is the picture in question:

In the proof of $$\lim_{x\to 0}\frac {\sin x}{x} = 1$$ involving the unit circle, how did they get the height of the largest right triangle to be $\tan(x)$? Shouldn't it be equivalent to $\sin(x)$? I feel as if the answer to this question is really obvious, but I've tried solving it by substitution of opposite, adjacent, etc. and it just doesn't work out.

Thanks!

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As the angle goes up to a right angle, notice that the thing in the picture labeled $\sin x$ goes up only as far as $1$, whereas the thing labeled $\tan x$ goes up to $\infty$. That should tell you that it certainly cannot be $\sin x$. See my answer below. –  Michael Hardy Apr 1 '14 at 15:36

If the radius of the circle is 1, then the triangle involving $\tan(x)$ is one whose base is 1 and whose height is some number, say $h$, and whose angle is given by $x$.

In such a case, the "adjacent" side has length 1, and the "opposite" side has length $h$. Thus

$$\tan(x) = \frac{\text{opposite}}{\text{adjacent}} = \frac{h}{1} = h$$

so the unknown height is given by $\tan(x)$ as claimed.

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But if you do the same with $sin x$ you get opposite over hypotenuse, hypotenuse is 1, so you get opposite. That's the same as $tan x$ . –  OpieDopee Apr 1 '14 at 15:42
@Ethan No, hypotenuse is not 1 for the triangle $ABD$ in question. The hypotenuse is $AD$, which is not $1$. –  6005 Apr 1 '14 at 16:46
@Goos: I'm referring to the hypotenuse AC. The hypotenuse for the triangle with hypotenuse AC is 1, making both $tan x$ and $sin x$ equal to opposite. –  OpieDopee Apr 1 '14 at 18:02
@Ethan Yes, $\tan x$ and $\sin x$ are both equal to opposite in different cases. "opposite" means a different thing depending on the triangle. But if the angle $x$ is the same, then no matter what opposite and adjacent are, $\frac{\text{opposite}}{\text{adjacent}}$ will always be the same value, and we call that $\tan x$. –  6005 Apr 1 '14 at 18:28

Hint: if you look at the two similars triangles $\hat{ACH}$ and $\hat{ADB}$, you have $$\frac{CH}{AH}=\frac{DB}{AB}$$ $$\frac{\sin(x)}{\cos(x)}=\tan(x)=DB$$ denoting $H$ the projection of $C$ on $AB$.

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As the angle goes up to a right angle, notice that the thing in the picture labeled $\sin x$ goes up only as far as $1$, whereas the thing labeled $\tan x$ goes up to $\infty$. That should tell you that it certainly cannot be $\sin x$.

It's the tangent because $\tan$ is opposite over adjacent, and the length of the adjacent side is $1$.

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Call the height $H$. Then $\tan x = {\rm opp}/{\rm adj} = H/1 = H$.

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