How can one prove the statement
limx→0sinxx=1
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 △ABC is 12sin(x). The area of the colored wedge is 12x, and the area of △ABD is 12tan(x). By inclusion, we get
12tan(x)≥12x≥12sin(x)
Dividing (1) by 12sin(x) and taking reciprocals, we get
cos(x)≤sin(x)x≤1
Since sin(x)x and cos(x) are even functions, (2) is valid for any non-zero x between −π2 and π2. Furthermore, since cos(x) is continuous near 0 and cos(0)=1, we get that
limx→0sin(x)x=1
Also, dividing (2) by cos(x), we get that
1≤tan(x)x≤sec(x)
Since sec(x) is continuous near 0 and sec(0)=1, we get that
limx→0tan(x)x=1
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