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Define $f\left(x\right)=\frac{\cos x}{x}$ $ f:\mathbb{R\backslash}\left\{ 0\right\} \longrightarrow\mathbb{R}$

So I need to determine if there is a continuous extension to $f\left(x\right)$ at a=0 and then find it if so.

What I thought of is since ${\displaystyle \lim_{x\to a^{-}}f\left(x\right)=-\infty}$ and $ {\displaystyle \lim_{x\to a^{+}}f\left(x\right)=\infty}$ , then it's not possible that there is a continuous extension to $f\left(x\right)$

Is that argument valid?

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Yes, you are right! This cases are called "essential singularities". –  emiliocba Jan 23 '12 at 12:16
    
@emiliocba: I'm not sure what you mean. Viewed as a function of a complex variable, $f$ has a pole at zero, not an essential singularity. (Remember that on the Riemann sphere there is only one point at infinity.) –  Pete L. Clark Jan 23 '12 at 12:27

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Yes, your argument is valid.

In fact, even half of your argument is valid: $\lim_{x \rightarrow a^+} f(x) = \infty$ is already enough to see that $f$ cannot extend to a continuous function $f: \mathbb{R} \rightarrow \mathbb{R}$.

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I'm not sure how to formalize that explantation, by intuition it's quite obvious why the function cannot be extend. –  sony jimbo Jan 23 '12 at 12:35
    
@sony jimbo: If $f$ could be defined at zero so as to be continuous, then $\lim_{x \rightarrow 0} f(x)$ must exist. But since $\lim_{x \rightarrow 0^+} f(x) = \infty$, this limit does not exist. –  Pete L. Clark Jan 23 '12 at 13:09

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