let f be differentiable at x=a.
Find lim (f(a+h)-f(a-h))/2h
     h goes to 0

how do you interpret these quotients in terms of slopes of the secant
lines to the graph of f

Hello again fmunshi!
We have that f(x) is differentiable at x=a.

"A real function is said to be differentiable at a point if its
derivative exists at that point..."

Differentiable --- from MathWorld
http://mathworld.wolfram.com/Differentiable.html

So, what is the derivative? Here's the definition, again from
MathWorld:

Derivative --- from MathWorld
http://mathworld.wolfram.com/Derivative.html

Go o to equation (6) in that page, to find that the derivative exists
at a point x if

lim     (f(x+h) - f(x))/h     exists  
h -> 0

Therefore, the fact that your funtion f is differentiable at a implies
that:

lim      (f(a+h)-f(a))/h     exists
h -> 0

Let's call this limit f'(a).

Now, you have to find 

lim      (f(a+h)-f(a-h))/2h 
h -> 0 

Let's take the term (f(a+h)-f(a-h))/2h and rewrite it:

(f(a+h)-f(a-h))/2h
=(f(a+h)-f(a-h)+f(a)-f(a))/2h
=1/2 (f(a+h)-f(a)+f(a)-f(a-h))/h
=1/2 (f(a+h)-f(a))/h + (f(a)-f(a-h))/h

Therefore, the problem has become to find

lim   1/2 (f(a+h)-f(a))/h + (f(a)-f(a-h))/h
h ->0

Now we can take the 1/2 out of the lim and we can use the fact that
lim(f(x)+g(x) = lim(f(x)) + lim(g(x)) to get

1/2 *[ lim  (f(a+h)-f(a))/h + lim (f(a)-f(a-h))/h  ]
       h->0                   h->0

But since h->0 and the derivative exists, then we know that these two
terms are equal. In particular, as we have defined above, they are
equal to f'(a).

Therefore, we find that the whole expression gives:

1/2 * [f'(a) + f'(a)]
=1/2 * [2*f'(a)]
=f'(a)

That's the solution to your question. How does this relate to the
slopes of the secant lines. Well, basically, both terms:

f(a+h)-f(a))/h and  (f(a)-f(a-h))/h

are the slopes of secant lines. The first one is the slope of the
secant line between points a+h and a. The second one is the slope of
the secant line between a and a-h. But we know that as h tends to 0,
these two secant lines tend to the slope of the tangent line of f at
point a, because f is derivable at a. They tend to exactly ther same
value f'(a), which is the slope of the tangent line to f at point a.
Check this page that illustrates the relationship between the slope of
secant lines and the slope of tangent lines.

Secant Line
http://www.ies.co.jp/math/java/calc/limsec/limsec.html

I hope this explanation was clear enough. If you have any doubts,
don't hesitate to request a clarification. Otherwise, I await your
rating and final comments.

Google search terms used
differentiable
secant line
tangent line

Best regards,
elmarto