How to make a loop until a function has a nonzero limit?

I'm making an application for L'hopitals rule so the condition has to be until one of the functions has derived enough to have a nonzero limit. For this application, I have to use a while loop so here's what I got already.

while (limit(f,x,0)==0 limit(g,x,0)==0)

a=diff(f(x));

b=diff(g(x));

end;

y=(limit(f,x,0))/(limit(g,x,0))

My problem is since I need to assign variables to differentiate the two functions and the variables cannot be the same as the ones I assigned to the functions, how can I tell the while loop to continue to derive the functions when the derived functions has a different variable?

Answer 1

[Matt J]

If you're just trying to avoid overwriting f and g, why not just make prior copies of them:

a=f;
b=g;

while (limit(a,x,0)==0 limit(b,x,0)==0)

a=diff(a(x));

b=diff(b(x));

end;

y=(limit(a,x,0))/(limit(b,x,0))

[Andy]

well I can't overwrite any variable according to the errors I'm getting

[Matt J]

Show the errors (by COPY/PASTE only)

Answer 2

[Star Strider]

I am not certain I completely understand your problem, but if you want to continue to differentiate your functions until you get a finite result, I suggest this approach:

syms a b c d f g x 

f(x) = sin(x)
g(x) = 1 - exp(x)

c = f(x)
d = g(x)
k1 = 1;
while (limit(c,x,0)==0) && (limit(d,x,0)==0)

      a=diff(c(k1,:));
      b=diff(d(k1,:));

      k1 = k1 + 1
      c(k1,:) = a
      d(k1,:) = b

end;

y = (limit(c(k1,:),x,0))/(limit(d(k1,:),x,0))

Vectors c and d store the intermediate results, so you do not overwrite your original functions.

[Andy]

I forgot to mention but the f and g in this function are function handles

[Star Strider]

I could only get this to work in the Symbolic Math Toolbox. I assumed that when you referred to limit in your code snippet, you were using it.

I just now tried the code I posted with f and g redefined as:

f = @(x) sin(x)
g = @(x) cos(x) - exp(x)

and it worked just as well as with the original function definitions.

NOTE: I'm using 2012b.