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Is there a difference between $(fg)(x)$ and $f(g(x))$? From my lecture slides

$\lim\limits_{x\to a} (f(x)+g(x)) = \lim\limits_{x\to a} f(x) + \lim\limits_{x\to a} g(x)$

Then in the next slide, it shows

$\lim\limits_{x\to a} g(f(x)) = g(\lim\limits_{x\to a} f(x))$

It seem to contradict. Shouldn't it be

$\lim\limits_{x\to a} g(f(x)) = \lim\limits_{x\to a} (gf)(x) = \lim\limits_{x\to a} g(x) \times \lim\limits_{x\to a} f(x)$

Or does $(gf)(x)$ mean $g(x)\cdot f(x)$ not $g(f(x))$? I thought it meant the later.

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Quick example: If $g(x)=x^3$ and $f(x)=x^2$, then $g(f(x))$ means $g(x^2)=(x^2)^3=x^2x^2x^2=x^6$, while $(gf)(x)=g(x)f(x)=x^3x^2=x^5$. The first is composition, not multiplication. – Jonas Meyer Jan 29 '12 at 8:27
Well, $g(f(x))$ is definitely composition, but whether $gf$ is multiplication or composition is a question of what convention has been agreed to by the parties to the exposition. – Gerry Myerson Jan 29 '12 at 8:31
@jiewmeng: I think you have come some way from posting scanned images of your work to typesetting it using TeX; keep it up. I just wanted to give a tip that would be hopefully useful: limits can be typeset as \lim_{x \to a} command (rather than \underset{x \to a} lim). This has multiple benefits: (1) This is simpler than what you wrote. (2) The lim symbol is now upright, and not in italics. (3) It automatically takes care of some spacing issues (I believe). – Srivatsan Jan 29 '12 at 8:37
@Srivatsan, its written not scanned :) didn't know I can scan into LaTeX. Thanks for the tip – Jiew Meng Jan 29 '12 at 8:42
up vote 3 down vote accepted

As noted multiple times already, it is not clear whether $gf$ is intended to mean the pointwise multiplication $(gf)(x) = g(x) \cdot f(x)$, or the composition $(gf)(x) = g(f(x))$. However irrespective of which convention you follow, your conclusion $$\lim\limits_{x\to a} g(f(x)) = \lim\limits_{x\to a} (gf)(x) = \lim\limits_{x\to a} g(x) \times \lim\limits_{x\to a} f(x)$$ is plain wrong. This is because,

  • in the first step $\lim\limits_{x\to a} g(f(x)) = \lim\limits_{x\to a} (gf)(x) $, you are using $gf$ to mean the composition $g \circ f$; whereas

  • in the very next step $\lim\limits_{x\to a} (gf)(x) = \lim\limits_{x\to a} g(x) \times \lim\limits_{x\to a} f(x)$, you follow the opposite convention that $gf$ is the pointwise multiplication of $g$ and $f$.

Thus the real problem is using a single notation to mean two different things simultaneously. And the fix is simple enough: pick one of the above conventions and apply it consistently.

By the way, the correct identities you are looking for are:

  • Composition rule: If $g$ is continuous at $\lim \limits_{x \to a} f(x)$, then $\lim\limits_{x \to a} g(f(x)) = g \left( \lim \limits_{x \to a} f(x) \right)$.

  • Multiplication rule: $\lim\limits_{x \to a} \left( g(x) \cdot f(x) \right) = \lim \limits_{x \to a} g(x) \cdot \lim \limits_{x \to a} f(x)$.

Note that I purposely avoided using the $gf$ notation.

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I have seen both conventions used in different contexts. Namely, in some texts, $gf$ stands for the function that maps $x$ to the product of $g(x)$ and $f(x)$, and in other texts $gf$ stands for the composition of $g$ and $f$, also known as $g\;{\scriptstyle \circ}\;f$.

Nonetheless, in my experience at least, the former usage is the more common of the two, especially in contexts where "pointwise" expressions like $f(x), g(x)$, $\lim\limits_{x\to a} g(f(x))$, etc. are being used (rather than those in which functions $f$, $g$, etc. are treated more abstractly, e.g. as morphisms in a category, without making reference their values at individual points). Therefore, in such a context a "pointwise" assertion like your proposed $\lim\limits_{x\to a} g(f(x)) = \lim\limits_{x\to a} (gf)(x)$, with its implicit use of the definition $gf \equiv g\;{\scriptstyle\circ}\;f$, would strike me as unconventional, and regrettably so, since it can so easily lead to confusion.

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