Question

Lim x-> 0

2x(cot3x)(sec4x)

I got to
2x(cos3x/sin3x)(1/sin4x)

But I'm not sure if..
cos x/x =1
I know sin x/x =1 and x/ sin x =1

could someone help me with this problem?

Answer 1

cos(x)/x -> inf as x->0

Answer 2

hmm.. okay.. can someone help me solve the problem?

Answer 3

did you rewrite every thing in terms of sines and cosines?

Answer 4

yes ^^^ it's up there

Answer 5

2x(cos3x/sin3x)(1/sin4x)

Answer 6

oops mistake
secant is the reciprocal of COSINE

Answer 7

you should get
\[\frac{\cos(3x)}{\cos(4x)}\times \frac{2x}{\sin(3x)}\]

Answer 8

why would I get that 2 up there?

Answer 9

sec = 1/x.. right? x is cosine

Answer 10

it is the \(2x\) in your question

Answer 11

\(x\) is not cosine. cosine is cosine, \(x\) is \(x\)

Answer 12

yes,, but why would it only go there?? ?

Answer 13

you mean instead of where?

Answer 14

2x(cos3x/sin3x)(1/sin4x)

it's strange that it skipped everything else and just went over sin??

Answer 15

or cos** I made a mistake there

Answer 16

you have
\[2x\cot(3x)\sec(4x)=2x\frac{\cos(3x)}{\sin(3x)}\times \frac{1}{\cos(4x)}\]
\[=\frac{2x\cos(3x)}{\sin(3x)\cos(4x)}\]
\[=\frac{\cos(3x)}{\cos(4x)}\frac{2x}{\sin(3x)}\]

Answer 17

i just arranged it so i can take limits i know
\(\cos(0)=1\) so we can ignore the first part, and take
\[\lim_{x\to 0}\frac{2x}{\sin(3x)}=\frac{2}{3}\]

Answer 18

is cos(0) always 1?

Answer 19

?

Answer 20

so cosx = 1?

Answer 21

cosine of 0 is 1, cosine of x depends on x

Answer 22

OH okay but in a limit where x-> 0, this will be true?

Answer 23

cosine is a function right? so for example if i have \(f(x)=x^2\) then \(f(1)=1\) but \(f\) of some other number is different

Answer 24

what would at lim x-> 0

x/cos x-1 be?

Answer 25

cosine is a continuous function, so the limit as x goes to 0 of cosine is \(\cos(0)=1\)

Answer 26

btw you just made be from ??? to OH right now. thank you

Answer 27

that is a different story
if you replace \(x\) by \(0\)
in that case you get
\[\frac{0}{0}\]so you have to do more work

Answer 28

right, okay thank you so much!

Answer 29

yw, by the way the answer to
\[\lim_{x\to 0}\frac{x}{\cos(x)-1}\] is it does not exist

Answer 30

oh okay, thank you I was just wondering what would happen there !