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The limit as x approaches zero of sin(x) over x can be determined using the squeeze theorem.
For 0 < x < pi/2, sin(x) < x < tan(x)
Divide by sin(x), and you get 1 < x/sin(x) < tan(x)/sin(x)
That is the same as 1 < x/sin(x) < 1/cos(x)
But the limit as x approaches zero of 1/cos(x) is 1,
so 1 < x/sin(x) < 1
which means that the limit as x approaches zero of x over sin(x) is 1, and that also means the inverse; the limit as x approaches zero of sin(x) over x is 1.
You can also solve this using deriviatives...
The deriviative d/dxx is 1, at all points. The deriviative d/dxsin(x) at x=0 is also 1.
This means you have the division of two functions, sin(x) and x, at a point where their slope is the same, so the limit reduces to 1 over 1, which is 1.
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