Question

Using L'Hopital's rule,Evaluate the limit.
lim as x approaches pi/4 9 tan u - 9 cot u divided by 4u-pi

Answer 1

Wouldn't this be the limit as u approaches pi/4?

Answer 2

still need this?

Answer 3

yes

Answer 4

do you know what l'hopital's rule is? ^_^

Answer 5

yes.. If the problem goes to 0/0 or infinity / infinity you take the derivative of both the numerator and denominator..

Answer 6

yes. so we have:

\[9 \lim_{u -> \pi/4}\frac{ \sec^2(u) + \csc^2(u) }{ 4 } = 9 *\frac{ \frac{ 1 }{ (\cos(\pi/4))^2 } +\frac{ 1 }{ (\sin(\pi/4))^2 } }{ 4 } = 9*\frac{ 2+2 }{ 4 } = 9\]

Answer 7

can explain to me step by step on how you did this?

Answer 8

can you help me with another problem?

Answer 9

we started with
\[9 \lim_{u ->\pi/4}\frac{ \tan(u) + \cot(u) }{ 4u - \pi }\]since this is 0/0 we take the derivative of them respectively [not to be confused with quotient rule]
derivative of tanu = sec^2(u)
derivative of cotu = -csc^2(u)
derivative of denominator is 4

and sec^2(u) = (sec(u))^2

and sec(u) = 1/cos(u)
csc(u) = 1/sin(u).

i used a calculator for sin and cos of pi/4 [even though don't have to]

Answer 10

i have to eat but if nobody helps by the time im done (like 10 minutes) i will

Answer 11

okay .. Thanks..

Answer 12

...

Answer 13

I posted the other question. .

Answer 14

limx→0 14csc^2(x)−14cot^2(x) using L'Hopital's rule..