limit x tends to 0 sin x^0/x
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1 Answer

If your question is lim x->0 sinx/x
Then solution is
Taylor expansion of sin(x) about x=0 is

sin(x) = x - x^3 / 3! + x^5 / 5! - x^7 / 7! + ...

So sin(x) / x = 1 - x^2 / 3! + ...

In the limit that x goes to zero, [sin(x)/x] = 1 - (0)^2 / 3! + ... = 1


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