hopital rule

limit(x to pie/2) -(x - pie/2) tanx

the limit is +1 but I don't get the answer. I applied product rule once but can't reach the answer.

In order to apply the L'Hopital rule you must have a quotient of 2 functions that tend both to 0. So what you do is transform the expression accordingly:

lim(x -> Pi/2) -(x - Pi/2)tan(x) = lim(x -> Pi/2) -(x - Pi/2) / cotan(x)

Now both the numerator and the denominator tend to 0 as x tends to Pi/2 so you apply the rule:

lim(x -> Pi/2) -(x - Pi/2) / cotan(x) = lim(x -> Pi/2) [-(x - Pi/2)]' / [cotan(x)]' = lim(x -> Pi/2) -1 / (-1/sin²x) = lim(x -> Pi/2) sin²x = 1

Thx.
Why is this rule invalid if either of the functions do not tend to zero?

And how to you check that the function tend to zero? you just equate to zero?

There are loads of sites about this rule. For example, you can take alook at the wikipedia,

http://en.wikipedia.org/wiki/L'H%C3%B4pital's_rule

which includes a proof where you'll see right away that both functions must tend to zero (btw, it works also when they both tend to infinity).

In order to calculate lim (x->h) f(x) , you first try to substitute the value h. If f(h) exists and is not undetermined, then this is the limit. Sometimes the subtitution leads to infinity which is acceptable as an answer, e.g. lim(x->3) 1/(x-3) or it may lead to an undeterminate expression, for eample:

lim(x->2) (x² - 4) / (x - 2)

If you substitute x=2 you get 0/0, so in this case you can't calculate the limit by substitution. But x² - 4 = (x + 2)(x - 2) so the limit becomes

lim(x->2) (x + 2)(x - 2) / (x - 2) = lim(x->2) (x + 2) = 4

Other examples of undetermined expressions are

x^1/(x-1) when x->1

(x + 3) / (3x - 5) when x-> infinite

sin(x) / x when x-> 0

etc.

thx mate.

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Originally Posted by yan03

thx mate.

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You're welcome (& you're welcome to rate my post, too ;) )

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Why is this rule invalid if either of the functions do not tend to zero?

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In any other case, you can easily determine what the limit should be.

Numerator and denominator can be a constant (c), 0, or infinity (i).

Cases: c/c, c/0, c/i; 0/c, 0/0, 0/i; i/c, i/0, i/i.

c/c = 1 always from basic algebra.
c/0: as you divide a constant by a smaller and smaller number, your result gets larger and larger. So, c/0 goes to infinity.
c/i: as you divide a constant by a larger and larger number, your result gets smaller and smaller. So, c/i goes to 0.

0/c = 0 from basic algebra.
0/0 = ?? from basic algebra. As you divide a smaller and smaller number by a smaller and smaller number, what do you get? It depends on how quickly both numbers get small--hence, you have to work with the numerator and denominator's derivatives and apply L'Hopital's rule.
0/i = 0. Both dividing a smaller and smaller number and dividing by a larger and larger number give you a smaller result. In the extreme, they both give you 0.

i/c = i. Eventually, dividing a larger and larger number by a constant does a meaningless amount of reduction to your final answer.
i/0 = i. Both dividing by a smaller and smaller number and dividing a larger and larger number give you a larger result. In the extreme, they both give you i.
i/i = ?? from basic algebra. As you divide a larger and larger number by a larger and larger number, what do you get? It depends on how quickly both numbers get large--hence, you have to work with the numerator and denominator's derivates and apply L'Hopital's rule.

Other indeterminate forms that L'Hopital's rule works for:
i*0 = i/c * c/i = i/i
i-i = ln(e^(i-i)) = ln(e^i/(e^i)) = i*ln(e/e) = i*ln(1) = i*0 [which is equivalent to i/i]
1^i = e^ln(1^i) = e^(i ln(1)) = e^(i*0) [i*0 is an indeterminate form for which the rule can be applied]
0^0 = e^(ln(0^0)) = e^(0*ln(0)) = (e^0)^ln(0) = 1^-i = 1/(1^i) [1^i is an indeterminate form for which the rule can be applied]
i^0 = e^(ln(i^0)) = e^(0*ln(i)) = e^(i*0) [i*0 is an indeterminate form for which the rule can be applied]

Edit: Added other indeterminate forms

Hope this helps to explain why it only works in certain cases.