How can one prove the statement
lim
without using the Taylor series of \sin, \cos and \tan? Best would be a geometrical solution.
This is homework. In my math class, we are about to prove that \sin is continuous. We found out, that proving the above statement is enough for proving the continuity of \sin, but I can't find out how. Any help is appreciated.
Answer
The area of \triangle ABC is \frac{1}{2}\sin(x). The area of the colored wedge is \frac{1}{2}x, and the area of \triangle ABD is \frac{1}{2}\tan(x). By inclusion, we get
\frac{1}{2}\tan(x)\ge\frac{1}{2}x\ge\frac{1}{2}\sin(x)\tag{1}
Dividing (1) by \frac{1}{2}\sin(x) and taking reciprocals, we get
\cos(x)\le\frac{\sin(x)}{x}\le1\tag{2}
Since \frac{\sin(x)}{x} and \cos(x) are even functions, (2) is valid for any non-zero x between -\frac{\pi}{2} and \frac{\pi}{2}. Furthermore, since \cos(x) is continuous near 0 and \cos(0) = 1, we get that
\lim_{x\to0}\frac{\sin(x)}{x}=1\tag{3}
Also, dividing (2) by \cos(x), we get that
1\le\frac{\tan(x)}{x}\le\sec(x)\tag{4}
Since \sec(x) is continuous near 0 and \sec(0) = 1, we get that
\lim_{x\to0}\frac{\tan(x)}{x}=1\tag{5}
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