Question #05bee

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Answer 1 · 2016-08-09T15:18:14

$lim_(xrarr0)(sin(x^2-x))/x = -1$ (I hope I've interpreted the question correctly.)

Use the fundamental trigonometric limit:

$lim_(theta rarr0) sin theta/theta = 1$

We'll make $theta = x^2-x$ by multiplying by one to get $x^2-x$ in the denominator.

$(sin(x^2-x))/x * (x-1)/(x-1) = (sin(x^2-x))/(x^2-x) * (x-1)$

Now the limit of the product is the product of the limits, so

$lim_(xrarr0)(sin(x^2-x))/x = lim_(xrarr0) (sin(x^2-x))/(x^2-x) * lim_(xrarr0)(x-1)$

$= (1)*(-1) = -1$