Question

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

Answer 1

-1

Explanation:

`color(blue)(1^(st) " Method")`

First we check if we can use L'Hôpital's rule. For this to be viable we need the limit to be of indeterminate form.

`(sin(0))/(2(0)^2 - 0) = 0/0`

So limit is indeterminate and we can use L'Hôpital's.

`lim_(xrarr0)(sinx)/(2x^2-x) = lim_(xrarr0)(d/(dx)(sinx))/(d/(dx)(2x^2-x))`

`=lim_(xrarr0)(cosx)/(4x-1) = (cos(0))/(4(0)-1) = 1/(-1) = -1`

`color(blue)(2^(nd) " Method")`

Rewrite limit as

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

It's a logical fallacy to use L'Hôpital's on the first term so we use alternative methods such as the Squeeze Theorem to show that

`lim_(xrarr0)(sinx)/x = 1`

So overall limit becomes:

`lim_(xrarr0)(sinx)/(2x^2-x) = lim_(xrarr0)(1)/(2x-1) = 1/(2(0) - 1) = 1/(-1) = -1`