Question

lim tan^2 3x/1-cosx

Answer 1

is it \[\lim_{x \rightarrow 0}\frac{ \tan^2(3x) }{1-cosx}??????????\]

Answer 2

x->0

Answer 3

how can i resolve this?

Answer 4

in these type of problems
we always first apply limit
put x=0
after applying limits, we get 0/0 form which is not possible
so we will apply L hospital rule
do u know how to apply L hospital rule?

Answer 5

no

Answer 6

in order to apply L hospital rule , we differentiate numerator and denominator separately
that is
\[\lim_{x \rightarrow 0}\frac{ \frac{ d }{ dx }(\tan^23x) }{ \frac{ d }{ dx }(1-cosx) }\]
now what will u get after differentiation?

Answer 7

i need apply trigonometric identities

Answer 8

can i use identi

Answer 9

According to me, with identities it will become complex

Answer 10

but is there solution using trigonometries?

Answer 11

may be
i think we need other people opinion

Answer 12

@whpalmer4

Answer 13

how is the solutuin using H hospital law?

Answer 14

Try doing L'Hospital twice...

Answer 15

i don,t learn yet

Answer 16

yeah Lhospital will be applied twice
i was going thru step by step explanation
but he wanted to go for solution using trigonometric identity

Answer 17

i multiply both up and down for 1+cos x

Answer 18

seems u want to go for trigonometric identities
show ur attempt first
i feel u r making this problem a very complex one by using trigonometric identites

Answer 19

why?

Answer 20

Well, feel free to demonstrate that it is not complex with your solution using trigonometric identities :-)

Answer 21

yah feel free buddy

Answer 22

i dont know l hospital law yet. the problem asking me to resolve using identities trigonometrics. i resolve other question bus this i cant. if you like i send you a file whith the resolution that i could resolve i will show yu, probably l hospital is more easy but i dont know yet

Answer 23

please dont be borred with me i only asking help

Answer 24

I m not bored
Actually i m also looking for the solution using trigonometric identities now
but m not able to solve too

Answer 25

and you could send me the answer using the hospital law?

Answer 26

using L hospital rule
\[\frac{ 2\tan(3x)*\sec^2(3x)*3 }{ 0 - (-sinx) }\]
after we are still getting 0/0 form
so we are going to apply L hospital rule
\[\frac{ \frac{ d }{ dx }6tanxsec^2x }{\frac{ d }{ dx }sinx }\]

Answer 27

oops it should be tan3x (by mistake i have written tanx in numerator)

Answer 28

\[\frac{ 6\tan3x*2secx*secxtanx + 6 \sec^23x*3*\sec^2x }{cosx }\]
now apply limits
\[\frac{ 0+18 }{ 1 }\]
\[18\]

Answer 29

this is the method of doing this problem
I assume u know the differentiation as well as product rule

Answer 30

niksva thank you a lot.

Answer 31

:) cheers

Answer 32

Okay, I've got it via trig identities. Boy, is it ugly :-)

Answer 33

\[\tan(u+v) = \frac{\tan(u)+\tan(v)}{1-\tan(u)\tan(v)}\]\[\tan (2 x+1 x)=\frac{\tan (x)+\tan (2 x)}{1-\tan (x) \tan (2 x)}\]
\[\tan (2 x)=\frac{2 \tan (x)}{1-\tan ^2(x)}\]
\[\tan (3 x)=\frac{\frac{2 \tan (x)}{1-\tan ^2(x)}+\tan (x)}{1-\frac{2 \tan ^2(x)}{1-\tan ^2(x)}}\]
\[\tan^2(3x) = \frac{\left(\frac{2 \tan (x)}{1-\tan ^2(x)}+\tan (x)\right)^2}{\left(1-\frac{2 \tan ^2(x)}{1-\tan ^2(x)}\right)^2}\]
\[\frac{\tan^2(3x)}{1-\cos (x)} = \frac{\left(\frac{2 \tan (x)}{1-\tan ^2(x)}+\tan (x)\right)^2}{\left(1-\frac{2 \tan ^2(x)}{1-\tan ^2(x)}\right)^2(1-\cos(x))}\]Multiply by \(\large \frac{(1+\sin(x))}{(1+\sin(x))}\) to get
\[\frac{\tan^2(3x)}{1-\cos (x)} = \frac{\left(\frac{2 \tan (x)}{1-\tan ^2(x)}+\tan (x)\right)^2(1+\cos(x))}{\left(1-\frac{2 \tan ^2(x)}{1-\tan ^2(x)}\right)^2(1-\cos^2(x))}\]
\[\frac{\tan^2(3x)}{1-\cos (x)} = \frac{\left(\frac{2 \tan (x)}{1-\tan ^2(x)}+\tan (x)\right)^2(1+\cos(x))}{\left(1-\frac{2 \tan ^2(x)}{1-\tan ^2(x)}\right)^2\sin^2(x)}\]
Now expand the numerator:
\[\frac{\tan^2(3x)}{1-\cos (x)} =
\frac{
\left(\frac{4 \tan ^2(x)}{1-\tan ^2(x)}+\frac{4 \tan ^2(x)}{\left(1-\tan ^2(x)\right)^2}+\tan ^2(x)\right)
(1+\cos(x))}

{\left(1-\frac{2 \tan ^2(x)}{1-\tan ^2(x)}\right)^2\sin^2(x)}\]
Now we divide out the \(\sin^2(x)\) from the \(\tan^2(x)\) in the numerator of the numerator:
\[\frac{\tan^2(3x)}{1-\cos (x)} =
\frac{
\left(\frac{4 \cos^2(x)}{1-\tan ^2(x)}+\frac{4 \cos ^2(x)}{\left(1-\tan ^2(x)\right)^2}+\cos ^2(x)\right)
(1+\cos(x))}

{\left(1-\frac{2 \tan ^2(x)}{1-\tan ^2(x)}\right)^2}\]

Now you can evaluate that at \(x = 0\)!

Answer 34

Uh, sorry, I said multiply by \[\frac{(1+\sin(x))}{(1+\sin(x))}\]but it clearly should be \[\frac{(1+\cos(x))}{(1+\cos(x))}\]

Answer 35

\[\lim_{x\rightarrow 0} \frac{\tan^2(3x)}{1-\cos (x)} = \lim_{x\rightarrow 0}
\frac{
\left(\frac{4 \cos^2(x)}{1-\tan ^2(x)}+\frac{4 \cos ^2(x)}{\left(1-\tan ^2(x)\right)^2}+\cos ^2(x)\right)
(1+\cos(x))}

{\left(1-\frac{2 \tan ^2(x)}{1-\tan ^2(x)}\right)^2} \]\[\lim_{x\rightarrow 0}
\frac{
\left(\frac{4 \cos^2(x)}{1-\tan ^2(x)}+\frac{4 \cos ^2(x)}{\left(1-\tan ^2(x)\right)^2}+\cos ^2(x)\right)
(1+\cos(x))}

{\left(1-\frac{2 \tan ^2(x)}{1-\tan ^2(x)}\right)^2} \]\[=\frac{
\left(\frac{4 *1}{1-0}+\frac{4 *1}{\left(1-0\right)^2}+1\right)
(1+1)}

{\left(1-\frac{2*0}{1-0}\right)^2} \]\[=\frac{(4+4+1)(2)}{1} \]\[= 18\]

Answer 36

whpalmer this resolution is realy very ugly. Thank you a lot.

Answer 37

Sorry