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I am confused by the examples in calculus textbook where they factor a polynomial to find the limit. I don't understand how the limits of $\frac{1}{x-3}$ and $\frac{x+3}{x^2-9} $ are the same. The only thing the book I read did was factor the polynomial, but it doesn't explain why that is possible.

I have thought of this? Trying to work the problem backwards.

Example: $$\lim_{x\to 3} \frac{1}{x-3} $$ Now multiplying by one should not change anything (I hope).

$$ \lim_{x\to 3} \frac{1}{x-3} \times \frac{x+3}{x+3} = \lim_{x \to 3}\frac{x+3}{x^2-9}$$

Now doesn't the transformed limit also now obtain a point of discontinuity at $x=-3$

yiyi
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  • It does not make sense to speak of the continuity at $x=-3$ of the function $f:\mathbb{R}-{3,-3}\rightarrow \mathbb{R}$ that sends $x$ to $ \frac{x+3}{x^2-9}$ – Amr Sep 04 '13 at 23:21
  • @amr if you input -3 into the second one, you get a zero on the bottom and isn't that a big No No? – yiyi Sep 04 '13 at 23:24
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    I don't understand the question. The title asks about factoring a limit and I don't see how you've factored anything. – Michael Albanese Sep 04 '13 at 23:25
  • Do you find any of the answers below acceptable? If so, you should accept one. If not, could you let us know what it is you are yet to understand? – Michael Albanese Sep 27 '13 at 19:01
  • @MichaelAlbanese hard to choose which one is the best answer. – yiyi Sep 30 '13 at 18:38
  • @yiyi: If they all sufficiently address your concerns, it doesn't matter which one you choose. Just pick whichever one you think is best; it is your question, so you can choose the best answer by whatever criteria you like. – Michael Albanese Sep 30 '13 at 20:12

4 Answers4

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The key is that $\dfrac1{x-3},\dfrac{x+3}{(x+3)(x-3)}$ differ only in that there exists a removable singularity in the latter at $x=-3$. They are identical almost everywhere, except that point.

obataku
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Formally it is because limits commute with multiplication wherever allowed, so $\lim_{x \rightarrow 3}( \frac{1}{x-3} \times \frac{x+3}{x+3})= \lim _{x \rightarrow 3}(\frac{1}{x-3}) \times\lim_{x \rightarrow 3}(\frac{x+3}{x+3}) = \lim_{x \rightarrow 3} (\frac{1}{x-3})\times 1$.

Informally the functions $f(x) =\frac{1}{x-3}$ and $h(x) = \frac{1}{x-3} \times \frac{x+3}{x+3}$ are equal wherever both are defined, but $f(x)$ is defined at an additional point. At this point $h(-3) = \frac{1}{-6} \frac{0}{0}$ which is undefined. However limits are not concerned with whether a function is defined at the point where we take the limit. A limit only measures the value a function "gets really close to" And you should be able to see that $h(x)$ "gets really close to" $\frac{1}{-6}$ as we move towards $-3$.

Daron
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If you look at the definition (e.g., here: http://en.wikipedia.org/wiki/Limit_(mathematics)), you will see that $\mbox{lim}_{x \leftarrow c} f(x)$ does not depend on the value of $f(x)$ when $x = c$ and does not even require $f(x)$ to be defined when $x = c$.

Rob Arthan
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Suppose you're asked to find the limit

$$\lim_{x \to 3}\frac{x+3}{x^2 - 9}.$$

That is, as $x$ gets closer and closer to $3$, what does $\frac{x+3}{x^2 - 9}$ get closer to (if anything)? As you are only interested in what happens for $x$ close to $3$, you can consider $x$ to be in some small deleted interval around $0$; in fact, the precise definition does effectively do this. In particular, what is happening at $x = -3$ does not effect the above limit. With this in mind, note that

$$\lim_{x \to 3}\frac{x+3}{x^2-9} = \lim_{x\to 3}\frac{x+3}{(x-3)(x + 3)}.$$

As we are only considering $x$ near $3$, $x + 3 \neq 0$ so we can cancel the factors; that is

$$\lim_{x \to 3}\frac{x+3}{x^2-9} = \lim_{x\to 3}\frac{x+3}{(x-3)(x + 3)} = \lim_{x\to 3}\frac{1}{x-3}$$

but of course this final limit does not exist.

Michael Albanese
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