I missed the lesson in calculus class and I was wondering if someone could fill me in before I go back and fail the next test :/
Quote from: Shout on October 01, 2005, 09:01 PM
I missed the lesson in calculus class and I was wondering if someone could fill me in before I go back and fail the next test :/
indeterminate :P forms (http://www.tpub.com/math2/36.htm)
Basically you have to get it out of indeterminate form via factoring, L'Hopital's rule, or some other method before you can evaluate the limit.
Edit: I see that this page doesn't cover L'Hopital's rule. L'Hopitals rule says that if you a function f(x) and the limit of f(x) as x->n has the indeterminate form 0/0 or infinity/infinity, then the limit of f(x) as x->n is equal to the limit of (the derivative of the numerator divided by the derivative of the denominator).
Example:
find limit( (x + 3)/(2x + 1) ) as x goes to infinity:
Clearly we have a case of infinity/infinity if you try substitution.
Therefore, limit( (x + 3)/(2x + 1) ) x->inf == limit ( d/dx(x+3) / d/dx(2x+1) )
derivative of x+3 is 1;
derivate of 2x+1 is 2;
limit(1/2) x-> infinity = 1/2 => limit((x+3)/(2x+1)) x->inf is 1/2.