Question

Tell me if I am using the L'H'S rule correct.

Answer 1

Hah, i wish I could tell you U_U you teach me bruh >.>

Answer 2

\[\huge\color{red}{ \lim_{x \rightarrow 3} ~~\frac{x^2-9}{x-3} }\]

I know what you will tell me, factor out cancel x-3 on top and bottom, but I have to use L'H'S rule.

Answer 3

\[\huge\color{red}{ \lim_{x \rightarrow 3} ~~\frac{x^2-9}{x-3} }\]
\[\huge\color{red}{ \frac{(x^2-9)'}{(x-3)'} }\]

so far so good? correct?

Answer 4

now, \[\huge\color{red}{ (f-g)'= f'g-g'f }\]

Answer 5

This is very interesting. What do you suppose I make of it?

Answer 6

\[\color{red}{ \frac{(x^2-9)'}{(x-3)'} = \frac{2x \times 9~~~+~~~0 \times x^2}{(1 \times 3)~~~+~~~0 \times x}=\frac{2x \times 9}{(1 \times 3)} =\frac{2x \times 9}{(3)}}\]

what? no it's wrong then.

Answer 7

How would I use L'H'S here?

Answer 8

The rule is correct but you have not taken the derivatives correctly. They should be 2x and 1. You do not multiply at all.

Answer 9

(f - g)' = f' - g'

Answer 10

ranga, no.

Answer 11

Except to plugin to the resulting equation.

Answer 12

IS the derivative of a constant=0 right?

Answer 13

Yes.

Answer 14

You are using the product rule for a derivative when there is subtraction.

Answer 15

OH ^

Answer 16

Yeah, and how do I derive when there is a subtraction?

Answer 17

(f - g)' = f' - g'

Answer 18

Ok, sorry then my bad., okay ranga.

Answer 19

You take the derivatives of the lower and upper equations separately, which gives you lim as x goes to 3 2x. Plug 3 into that limit to get lim x goes to. 3=6

Answer 20

\[\frac{2x -0}{1-0}\]?

Answer 21

That is correct.

Answer 22

So, to get the final answer, I plug in 3 for x?

Answer 23

yes. lim x->3 (2x) = 2*3 = 6

Answer 24

after L'H'S of course )

Answer 25

Thank you ranga!

Answer 26

You are welcome.

Answer 27

Very good job gentlemen. I'm impressed.