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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 $1$"?

Paul
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Jim_CS
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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...) – David Mitra Feb 08 '12 at 00:14
  • It seems an odd question – Henry Feb 08 '12 at 00:15
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    @Henry, it's the kind of question calc teachers give to see whether students really understand limits and functional notation. – Gerry Myerson Feb 08 '12 at 00:34
  • 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. – Jim_CS Feb 08 '12 at 01:19
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    Relevant is Leibniz' law. $f(3)$ is indistinguishable from $9.$ Now what is $\lim\limits_{x \to 1} c$ for any constant function $c$? –  Feb 08 '12 at 04:19

2 Answers2

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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.

tomcuchta
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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 $b=9$.

rschwieb
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