L'Hospital's Rule.

Question

L'Hospital's Rule.

parthkohli · Answer

@hartnn I usually simplify limits by graphing or plugging in numbers or even simplifying the function. Can you teach me the L'Hospital's Rule?

parthkohli · Answer

I've heard that we differentiate the numerator and the denominator. Is there anything else to do?

hartnn · Answer

we differentiate provided the function is in the form of 0/0 or infinity/infinity when the variable is directly substituted its limit.
so,yes.First we need to check whether the function is in this form.

parthkohli · Answer

$$\lim_{x \to 3} {x^2 \over x - 3} $$I can apply the rule to this, right?

parthkohli · Answer

$$\lim_{x \to3} {2x \over 1} $$I took the derivative of the numerator and denominator.

hartnn · Answer

we can differentiate twice,thrice.....and so on,provided each time we get that form.
once we don't get it, we cannot use L'Hospitals.

parthkohli · Answer

$$\lim_{x \to 3} {x^2 \over x- 3} = 6x $$I got this.

parthkohli · Answer

Sorry, it's just 6.

hartnn · Answer

nopes,u cannot.the numerator does not go to 0.

parthkohli · Answer

Can you give me an example and solve?

hartnn · Answer

that limit is essentially infinity.

parthkohli · Answer

Okay, but can you set up a question and then solve it?

hartnn · Answer

the last problem u posted , the one with e^x
had 0/0 form ,differentiate it and tell me what u get.

hartnn · Answer

i mean differentiate numerator and denominator separately
u can us L'Hospital,because it had the form of 0/0

parthkohli · Answer

All right, a second please.

parthkohli · Answer

$$\lim_{x \to 0} \; \; {e^{x} -1 \over x}$$$$\lim _{x \to 0} \; \; {e^x \over 1} $$Is that correct?

parthkohli · Answer

$$\lim_{x \to 0} \; \; e^x  $$$$\implies 1$$

parthkohli · Answer

:)

parthkohli · Answer

Thank you very much! @hartnn

hartnn · Answer

thats correct :)
welcome.
just don't forget to check first,whether the form is 0/0 or infinity/infinity.

parthkohli · Answer

The form is $0 \over 0$

hartnn · Answer

yup,
an example of the other form
$$\lim_{x \rightarrow \infty}\frac{4x^2+3x+1}{2x^2+6x-9}$$
try this.

parthkohli · Answer

That's infinity/infinity. Let me try!

parthkohli · Answer

$$ \lim_{x \to \infty} \; \; {8x + 3  \over 4x + 6 }$$

parthkohli · Answer

Again differentiate?

parthkohli · Answer

$$\lim_{x\to\infty} \; \; {8 \over 4} $$$$\lim_{x \to \infty}  2$$$x$ isn't even there.

parthkohli · Answer

Is the answer just 2?

hartnn · Answer

yup,because the form is still same.so u can.

parthkohli · Answer

Then is the answer 2?

hartnn · Answer

yup,2.that simple.

parthkohli · Answer

Haha! Yes, it is very, very simple.

hartnn · Answer

did u just start with limits?know standard formulas?