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Let $ f(x) = x^2$

What is $\displaystyle\lim_{x \to 1}f(3)$

What is this statement saying in plain english?

Is it "What is $f(3)$ approaching as $x$ approaches $1$"?

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    Yes. But did you mean to write $\lim\limits_{x\rightarrow1} f(x)$? Here, the interpretation is "what is $f(x)$ approaching as $x$ approaches 1". (As written, the interpretation is correct, despite the fact that $f(3)$ is always 9. "Approaching" does not rule out being equal to when refering to the function you're taking the limit of...)2012-02-08
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    It seems an odd question2012-02-08
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    @Henry, it's the kind of question calc teachers give to see whether students really understand limits and functional notation.2012-02-08
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    It is a bit of an odd question alright lol, doing some math after a few pints. Just got a weird mental block about this statement.2012-02-08
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    Relevant is [Leibniz' law](http://en.wikipedia.org/wiki/Identity_of_indiscernibles). $f(3)$ is indistinguishable from $9.$ Now what is $\lim\limits_{x \to 1} c$ for any constant function $c$?2012-02-08

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Your interpretation is correct.

As written, it may help to think of this in the following manner: define $g(x) = f(3) = 9$, (i.e. $g$ is a constant function). Then

$$\displaystyle\lim_{x \rightarrow 1} f(3) = \displaystyle\lim_{x \rightarrow 1} g(x) = \displaystyle\lim_{x \rightarrow 1} 9.$$

Of course the value of the limit is $9$.

I think the "purpose" of this is to explain notation, but that's only a guess.

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$f(3) = 9$

So that the rule for limits that applies here is;

Limit of a constant, $b$:

$\lim_{x\rightarrow c} b = b$

where $c=1$ and $b=9$.