Finding the limit $\lim_{x\to-\infty} (2x)/(2x-1)^2$. Studying for a midterm:
Let $f(x)=\frac{2x}{(2x-1)^2}$
Then $\lim_{x\to-\infty} f(x)$ is:
Now keep in mind I'm shaky on how to do infinity limits.
I have $f(x)=\frac{2x}{(2x-1)^2}$
Remove x by dividing by the highest common denominator:
$=\frac{2+\frac1x}{(2-\frac1x)^2}$
$\frac1x$=$0$
so:
$=\frac{2+0}{(2-0)^2}$
$=\frac24$
$$ \lim_{x\to-\infty} f(x)\frac{2x}{(2x-1)^2}=\frac12$$
Although for some reason I don't think this is right. Since I feel like I'm finding the limit for a positive infinity function. I can't find help through other sources, so I would appreciate some help.

Your mistake is in the denominator and you should write
$$\frac{2x}{(2x-1)^2}=\frac{2x}{x^2(2-\frac{1}{x})^2}$$
and then you simplify and you pass to the limit to find the result $0$.
