# What is positive-0 squared minus positive-0?

I've got a basic limit problem that I think I'm solving the right way, but I've run into something that looks confusing enough to make me wonder if I'm doing it right.

$$\lim_{y\to0} \frac{1}{y^2-y} + \frac{1}{y}$$

I tried approaching this by looking at the Left Hand Limit and Right Hand Limit. Though when I plug in $0^+$, I get

$$\lim_{y\to0} \frac{1}{0^+-0^+} + \frac{1}{0^+}$$

What is $0^+-0^+$? Is it still positive? or is it just 0, or in this case, would it be negative? I'm leaning toward negative because for all $0\lt y\lt 1$, $y^2$ will be less than $y$, so then wouldn't I have an infinitely small negative number as a result?

And then, if that turns out to be true, it would appear that this problem is unsolvable by that method then, seeing as I'd have $+\infty + -\infty$, which is undefined.

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Clearly $y^2-y\to 0$ as $y\to 0$, as does $y$, so both fractions are blowing up, and their difference is an indeterminate form. Combine them over a common denominator; if you do the algebra correctly, you’ll be able to simplify the resulting fraction to one whose limit as $y\to 0$ is easy to compute.

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Oh, silly me - I never even noticed that I could factor out $y(y-1)$ and then combine the fractions and reduce it to $1/y-1$. It's been too long since I did basic algebra. :P – agent154 Sep 20 '12 at 3:01

Dividing by $0$ is nonsensical. And there is only one $0$.

Bring your expression to a common denominator and simplify.

Remark: You could get useful information from your calculator. Calculate $\dfrac{1}{y^2-y}+\dfrac{1}{y}$ for some values of $y$ not far from $0$, but of course not equal to $0$, like $y=0.001$ or $y=-0.002$.

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It's $-1$, silly! Obviously, you're very confused based on your argument. I suggest coming back at the problem with fresh eyes after you've taken a break from it.

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Look at your answer from the original poster's point of view. Is it helpful? – Rick Decker Sep 24 '12 at 20:51