Question

Csc / general trig limits question! Can anyone please help??

Answer 1

\[\lim_{x \rightarrow 0} \frac{ 2x \csc(2x) }{ \cos(5x) }\]

Answer 2

And if someone could walk me through the process, that would be awesome :/

Answer 3

do you know csc(x) = ?

Answer 4

how can you rewrite csc(x) ?

Answer 5

looke here please http://www.kkuniyuk.com/TrigBasicIDsAnyClass.pdf

Answer 6

\[\lim_{x \rightarrow 0}\frac{ 2x \csc 2x }{ \cos 5x }=\lim_{x \rightarrow 0}\frac{ 2x }{ \sin 2x \cos 5x }\]
\[=\lim_{x \rightarrow 0}\frac{ 1 }{ \frac{ \sin 2x }{ 2x } }\lim_{x \rightarrow 0}\frac{ 1 }{ \cos 5x }=?\]

Answer 7

So, csc can be re-written as \[\csc (x) =\frac{ 1 }{ \sin (x) }\] So, that start's the process of manipulating it? Do I use the 2x?

Answer 8

Hmmm, so the explanation.. \[\csc (x) = \frac{ 1 }{ \sin (x) }\]

Thus, this can be manipulated to: \[\frac{ 1 }{ \sin (2x) } \] ?? What about the other 2x?

Answer 9

(not focusing on cos yet)

Answer 10

sin(2x) = 2sinxcosx

Answer 11

cos(5x) looke it here please

http://2000clicks.com/mathhelp/GeometryTrigEquivCos3xEtcB.aspx

Answer 12

\[\lim_{x \rightarrow 0}=\lim_{2x \rightarrow 0}\]
\[\lim_{x \rightarrow 0}\frac{ \sin 2x }{ 2x }=1\]
\[\lim_{x \rightarrow 0}\cos 5x=1\]

Answer 13

So using the properties, they would both reduce to one and then 1 x 1 = 1 and thats the limit?

Answer 14

yes

Answer 15

correct.