Question

what is D.L.Hospital rule in Limits Topic?

Answer 1

you can apply L'hop... only when you have cases like:
\( \LARGE \frac{\infty}{\infty }\) and \(\LARGE \frac{0}{0}\) I call it "cheating" lol

Answer 2

What lol! I am just asking for ur help to make me understand this.

Answer 3

... hmm , understand what? ... you should know derivatives to apply L'H

Answer 4

ok let it go

Answer 5

can I provide a dizzy example? ... I'm not sure if you're asking for this :(

Answer 6

ya

Answer 7

Go ahead, all are waiting @Kreshnik

Answer 8

ex. Solve...
\[\Large \begin{array}{l}\mathop {\lim }\limits_{x \to 3} \frac{{{x^2} - 9}}{{x - 3}}\\\\\\\mathop {\lim }\limits_{x \to 3} \frac{{{x^2} - 9}}{{x - 3}} = \mathop {\lim }\limits_{x \to 3} \frac{{\left( {x - 3} \right)\left( {x + 3} \right)}}{{x - 3}} = \\\\ = \mathop {\lim }\limits_{x \to 3} \left( {x + 3} \right) = 3 + 3 = 6\\\\L'Hospital\\\\\mathop {\lim }\limits_{x \to 3} \frac{{{x^2} - 9}}{{x - 3}} = \mathop {\lim }\limits_{x \to 3} {\left( {\frac{{{x^2} - 9}}{{x - 3}}} \right)^'} = \\ = \mathop {\lim }\limits_{x \to 3} \frac{{2x}}{1} = 2 \cdot 3 = 6\end{array}\]

Answer 9

How did u write 2x in the last step?

Answer 10

... let me add more detail to L'H
\[\LARGE \left({x^2-9\over x-3}\right)^'=\left[{(x^2)-9'\over x'-3'}\right]=\]
\[\LARGE {2x^{2-1}-0\over 1-0}=2x=...\]

using chain rule:
\[\LARGE (n^u)'=u\cdot n^{u-1 }\]

Answer 11

\[\LARGE \left({x^2-9\over x-3}\right)^'=\left[{(x^2)'-9'\over x'-3'}\right]=\]

Answer 12

oh thanx kreshnik dude!.:(

Answer 13

Mahesh, just take the derivative of top and bottom. If it's still 0/0 you can do it again.

Answer 14

\[\lim_{x \rightarrow c}\frac{f(x)}{g(x)}=\lim_{x \rightarrow c}\frac{f'(x)}{g'(x)}\]Basically the rule says that the limit of one function divided by another function is the same as the limit of the first derivative of the numerator divided by the derivative of the denominator.

Answer 15

thanx Mertsj!

Answer 16

yw