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So the lecturer's assistant is saying that the following limit exists and that it is $\infty$

So the equation $$\lim_{x \to 2} \frac{2x+4}{x^2-4} $$

Now, if I go ahead and simplify the expression first I end up with

$$\lim_{x \to 2} \frac{2}{x-2} $$

and as far as I can tell this limit does not exist. Reason being that the Left Hand Side limit will be $-\infty$ and the Right Hand Side limit will be $+\infty$.

Can someone please verify if I'm in the wrong here?

Much appreciated.

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    $\begingroup$ It should simplify to $2/(x-2)$, and yes, the limit does not exist. $\endgroup$ Oct 22, 2013 at 7:41
  • $\begingroup$ Thank you, I'll update the simplified answer in question. $\endgroup$
    – Richard
    Oct 22, 2013 at 7:55

2 Answers 2

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$$\lim_{x\to 2}{\frac{2x+4}{x^2-4}} = \lim_{x \to 2}{\frac{2(x+2)}{(x-2)(x+2)}}$$ $$ \lim_{x\to 2}{\frac{2}{x-2}}$$ $$\text{Left-hand limit} = \lim_{x\to 2^-}{\frac{2}{x-2}} = -\infty$$ $$\text{Right-hand limit} = \lim_{x\to2^+}{\frac{2}{x-2}}= +\infty$$ $$\text{Left-hand limit} \ne \text{Right-hand limit}$$ Hence limit doesn't exist.

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It depends on the conventions used. Some traditions distinguish three cases:

\begin{align} &\lim_{x\to x_0}f(x)=+\infty \\ &\lim_{x\to x_0}f(x)=-\infty \\ &\lim_{x\to x_0}f(x)=\infty \end{align} where the last means $\lim_{x\to x_0}|f(x)|=+\infty$. According to this convention, one can say that $\lim_{x\to0}1/x=\infty$, which is essentially the same as your case.

Old textbooks in Italy had this three branched distinction, probably because of the influence from geometry: for rational functions “infinity without sign” actually has a geometric meaning. This isn't really useful for functions that are not rational and in more modern approaches “infinity without sign” isn't found any more. But traditions and habits (especially when they are proved faulty) are hard to die…

Ask the assistant what definitions he/she's using.

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