lim as x->0 of f(x) where f(x)= (1 /x^2)sin(x/2) i plugged in numbers close to zero and discovered the limit to be 0.5 but how do i solve this algebraically.
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Hello, Originally Posted by myoplex11 lim as x->0 of f(x) where f(x)= (1 /x^2)sin(x/2) i plugged in numbers close to zero and discovered the limit to be 0.5 but how do i solve this algebraically. If it is indeed (1/x²)sin(x/2), then the limit is not 0.5, it's infinite >.> If it is (1/x)sin(x/2), substitute t=x/2 (---> x=2t) and you'll find a limit you know. If it is (1/x²)sin(x²/2), substitute t=x²/2.
Originally Posted by Moo Hello, If it is indeed (1/x²)sin(x/2), then the limit is not 0.5, it's infinite >.> If it is (1/x)sin(x/2), substitute t=x/2 (---> x=2t) and you'll find a limit you know. If it is (1/x²)sin(x²/2), substitute t=x²/2. actually, in the first case, the limit doesn't exist. the one-sided limits are not equal
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