# 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 $0$.

• $\ddot\smile$ +1 Commented Oct 11, 2013 at 17:21

$$\frac{2x}{(2x-1)^2}=\frac{\frac2x}{\left(2-\frac1x\right)^2}\text{ and not}\frac{2+\frac1x}{\left(2-\frac1x\right)^2}$$

Alternatively putting $\frac1x=h$

as $x\to-\infty, h\to0^-$

So, $$\lim_{x\to-\infty}\frac{2x}{(2x-1)^2}=\lim_{h\to0^-}\frac{2h}{(2-h)^2}$$

Can you take it from here?

• So it's $\frac02$ meaning the limit is $-\infty$? Commented Oct 8, 2013 at 18:26
• @Unknown, $$\frac0a=?$$ if $a$ is non-zero finite number? Commented Oct 8, 2013 at 18:28
• $=0$ Easy enough, thanks very much for your help. Commented Oct 8, 2013 at 18:30