Question

LIMIT QUESTION

Answer 1

@UnkleRhaukus

Answer 2

\[\lim\limits_{x\rightarrow0}\frac{\sin(px)}{qx}=\]

Answer 3

@UnkleRhaukus What ?

Answer 4

is that your question?
or
\[\lim\limits_{x>0}\frac{\sin(px)}{qx}=\]
l'Hôpital's rule

Answer 5

No i my ques x approaches to zero

Answer 6

in*

Answer 7

ok so do you know l'Hôpital's rule ?

Answer 8

aND i Dont know what l'Hôpital's rule Is ? :(

Answer 9

Please Help Me @UnkleRhaukus

Answer 10

p/q is the answer

Answer 11

I need the explanation :(

Answer 12

the rule says that if your trying to evaluate the limit;
where simple substitution yields 0/0

differentiate the numerator and the denominator , and the limit will be equal

Answer 13

Provide Examples @UnkleRhaukus

Answer 14

just multiply numerator and denominator by p

Answer 15

multiply that fraction by pq/pq

Answer 16

u ll get ur answer urself

Answer 17

I think there are a few ways to do this one using the squeeze theorem. Unfortunately I can't remember the steps <:o

Khan Academy has a nice geometric proof if you care to look at that. :o

Answer 18

zepdrix sqeeze theoram is not needed here

Answer 19

Please Link me To it @zepdrix

Answer 20

http://www.khanacademy.org/math/precalculus/v/proof--lim--sin-x--x

Answer 21

If i follow @anup39

Answer 22

\[\lim\limits_{x\rightarrow0}\frac{\sin(px)}{qx} \quad \begin{array}{c}&\tiny{ l'H }\quad\\ &=\end{array}\lim\limits_{x\rightarrow0}\frac{\frac{\text d}{\text dx}\sin(px)}{\frac{\text d}{\text dx}qx} \]

Answer 23

What do i do next ? @UnkleRhaukus