Question

How do you find the limit of `(sin(x))/(2x²-x)` as x approaches 0?

Answer 1

Use algebra and `lim_(thetararr0)sintheta/theta = 1` to get a limit of `-1`

Explanation:

`lim_(xrarr0)(sin(x))/(2x²-x) = lim_(xrarr0)sinx/(x(2x-1))`

`= lim_(xrarr0)(sinx/x*1/(2x-1))`

`= lim_(xrarr0)sinx/x*lim_(xrarr0)1/(2x-1)`

`= 1*1/(0-1) = -1`