What's the point of $\eta$-conversion in lambda calculus?
Update [2011-09-20]: I expanded the paragraph about $\eta$-expansion and extensionality. Thanks to Anton Salikhmetov for pointing out a good reference.

$\eta$-conversion $(\lambda x . f x) = f$ is a special case of $\beta$- conversion only in the special case when $f$ is itself an abstraction, e.g., if $f = \lambda y . y y$ then $$(\lambda x . f x) = (\lambda x . (\lambda y . y y) x) =_\beta (\lambda x . x x) =_\alpha f.$$ But what if $f$ is a variable, or an application which does not reduce to an abstraction?

In a way $\eta$-rule is like a special kind of extensionality, but we have to be a bit careful about how that is stated. We can state extensionality as:


for all $\lambda$-terms $M$ and $N$, if $M x = N x$ then $M = N$, or
for all $f, g$ if $\forall x . f x = g x$ then $f = g$.


The first one is a meta-statement about the terms of the $\lambda$-calculus. In it $x$ appears as a formal variable, i.e., it is part of the $\lambda$-calculus. It can be proved from $\beta\eta$-rules, see for example Theorem 2.1.29 in "Lambda Calculus: its Syntax and Semantics" by Barendregt (1985). It can be understood as a statement about all the definable functions, i.e., those which are denotations of $\lambda$-terms.

The second statement is how mathematicians usually understand mathematical statements. The theory of $\lambda$-calculus describes a certain kind of structures, let us call them "$\lambda$-models". A $\lambda$-model might be uncountable, so there is no guarantee that every element of it corresponds to a $\lambda$-term (just like there are more real numbers than there are expressions describing reals). Extensionality then says: if we take any two things $f$ and $g$ in a $\lambda$-model, if $f x = g x$ for all $x$ in the model, then $f = g$. Now even if the model satisfies the $\eta$-rule, it need not satisfy extensionality in this sense. (Reference needed here, and I think we need to be careful how equality is interpreted.)

There are several ways in which we can motivate $\beta$- and $\eta$-conversions. I will randomly pick the category-theoretic one, disguised as $\lambda$-calculus, and someone else can explain other reasons.

Let us consider the typed $\lambda$-calculus (because it is less confusing, but more or less the same reasoning works for the untyped $\lambda$-calculus). One of the basic laws that should holds is the exponential law $$C^{A \times B} \cong (C^B)^A.$$ (I am using notations $A \to B$ and $B^A$ interchangably, picking whichever seems to look better.) What do the isomorphisms $i : C^{A \times B} \to (C^B)^A$ and $j : (C^B)^A \to C^{A \times B}$ look like, written in $\lambda$-calculus? Presumably they would be $$i = \lambda f : C^{A \times B} . \lambda a : A . \lambda b : B . f \langle a, b \rangle$$ and $$j = \lambda g : (C^B)^A . \lambda p : A \times B . g (\pi_1 p) (\pi_2 p).$$
A short calculation with a couple of $\beta$-reductions (including the $\beta$-reductions $\pi_1 \langle a, b \rangle = a$ and $\pi_2 \langle a, b \rangle = b$ for products) tells us that, for every $g : (C^B)^A$ we have $$i (j g) = \lambda a : A . \lambda b : B . g a b.$$
Since $i$ and $j$ are inverses of each other, we expect $i (j g) = g$, but to actually prove this we need to use $\eta$-reduction twice: $$i(j g) = (\lambda a : A . \lambda b : B . g a b) =_\eta (\lambda a : A . g a) =_\eta g.$$
So this is one reason for having $\eta$-reductions. Exercise: which $\eta$-rule is needed to show that $j (i f) = f$?