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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.

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2 Answers 2

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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 $0$.

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  • $\begingroup$ $\ddot\smile$ +1 $\endgroup$
    – Mikasa
    Commented Oct 11, 2013 at 17:21
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$$\frac{2x}{(2x-1)^2}=\frac{\frac2x}{\left(2-\frac1x\right)^2}\text{ and not}\frac{2+\frac1x}{\left(2-\frac1x\right)^2}$$

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?

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

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