Expert: Scotto - 5/3/2011

Question
QUESTION: what is L-hosptal rule and how it is used to evaluate limits of the form 0/0 ?plzzz explain clearly with examples.

ANSWER: Noy only is 0/0 involved, but infinity/infinity is involved as well.

L'hospital's rule states that if the top and bottom of a fraction both go to 0 of both go to infinity as x goes to 0 or x goes to infinity, the top and bottom can be differentiated.


For example, if we have (3x^2 + 5x + 4)/(6x^2 + 7x + 3) and are taking the limit as x goes to infinity, we get infinity/infinity.  Taking the derivative of the numerator gives 6x + 5
and taking the derivative of the denominator gives 12x + 7.

The limit(x->0)[(6x+5)/(12x+7)] is still 0/0, so differentiate the top and differentiate the bottom again.  This gives lim(*x->0)(6/12), and that is 1/2 since there is no x involved.


For another example, if we have (e^x - x - 1)/x^2 and are taking the limit as x->0,
the top is 1 - 0 - 1 = 0  and the bottom is 0^2 = 0.

This means we can differentiate the numerator and the denominator separately.
This gives e^x-1 for the top and 2x for the bottom.

If we then look at e^0 - 1 we get 0 and if we look at 2(0) we get 0,
so the fraction is still 0/0.

This means we can again separately differentiate the top and bottom.
This gives e^x/2, and at x-0 that is 1/2.

Thus we can say lim(x->0)[(e^x - x - 1)/(x^2) = 1/2.


---------- FOLLOW-UP ----------

QUESTION: Like there is chain rule for differentiation,is there any direct way for difficult problems in Integration??plzzzzzzzz give a clear solution

Answer
All problems can be differentiated by the power rule, exponential rule, or one of the trig rule using the chain rule.  However, not all problems can be integrated exactly.  Several of them are known if they can be integrated, and some are known to not be integrable, but there are still a few that we have no clue.  To solve some of them, the u-v method is used.  This is similar to the product rule in finding derivatives.

Differentiating is similiar to squaring an integer and getting an integer - it can be done to all of them.  Doing the inverse involes taking squareroots.  This can easily be shown to produce a non-integer.  In fact, it can be shown to produce a number with a non-repeating infinite length squareroot.

For exampe, 2^2 = 4, but squareroot(2) = 1.41421 35623 73095 04880 16887 24209 69807 85696 71875 37694 80731 76679 73799.... it has no end, and yet it seems so simple.

While I'm on that subject, let i = sqrt(-1).  It is known that e is irrational, pi is irrational, and i is not real, but it can be shown that e^(i*pi) = -1.
This is because the exponential of complex numbers can be expressed as a trig funciton.
That is, e^(a+bi) = e^a(cos b + i*sin b).  If a + bi is i, a=0 and b=pi.
Thus, e^(0+pi*i) = [e^0](cos(pi) + i*sin(pi)).  It is known that e^0 = 1, cos(pi) = -1,
and sin(pi) = 0, so that is 1(-1 + i(0)) = -1.