Question

calculate the limit. can someone please help me on how to do this?

Answer 1

\[\lim_{x \rightarrow 2\Pi} \frac{ \sin5x }{\sin3x }\]

Answer 2

@precal

Answer 3

\[\large \frac{ \sin 5x }{ \sin 3x }=\frac{ \sin5x }{ \sin 3x }\frac{ \frac{ 5 }{ 5x } }{\frac{ 3 }{ 3x } }=\frac{ 5\frac{ \sin 5x }{ 5x } }{ 3\frac{ \sin 3x }{ 3x } }\]

Answer 4

Perhaps this is easier to visualize:

\[\frac{ \sin5x }{ \sin3x } * \frac{ 5 }{ 5 } * \frac{ 3 }{ 3 } = \frac{ \sin5x }{ 5 }*\frac{ 3 }{ \sin3x }*\frac{ 5 }{ 3 }\]

Answer 5

i don't get it. is that all? @BluFoot @precal @sirm3d

Answer 6

let me modify my work, since x -> 2pi\[\large \frac{ \sin5x }{ \sin3x }=\frac{ \sin5(x-2\pi + 2\pi)}{ \sin 3(x-2\pi + 2\pi) }\]

Answer 7

applying the trigonometric identity to the sine of a sum of two angles, and simplifying\[\large \frac{\sin5(x-2\pi + 2\pi)}{\sin 3(x-2\pi + 2\pi)}=\frac{\sin5(x-2\pi)}{\sin 3(x-2\pi)}\]so we introduce\[\large x - 2\pi\]

Answer 8

\[\large \frac{ \sin 5(x-2\pi) }{ \sin 3(x-2\pi) }\frac{ 5 }{ 5 }\frac{ 3 }{ 3 }\frac{ x-2\pi }{ x-2\pi }\] we collect factors \[\large \frac{ \sin 5(x-2\pi) }{ 5(x-2\pi) }\frac{ 3(x-2\pi) }{ \sin 3(x-2\pi) }\frac{5}{3}\]

Answer 9

\[\large \lim_{x \rightarrow 2\pi} \frac{ \sin 5(x-2\pi) }{ 5(x-2\pi) }=1\]

Answer 10

we are using \[\large \lim_{x \rightarrow 0}\frac{ \sin x }{ x }=1\] and the more general form \[\large \lim_{x \rightarrow a}\frac{ \sin(x-a) }{ (x-a) }\]

Answer 11

but why did you do this? x−2π+2π

Answer 12

we need you use the general form above ^

Answer 13

where a = 2pi

Answer 14

yeah i get a = 2pi but why 2-2pi and then again +2pi?

Answer 15

Forget what he said, it's really confusing. Use the answer I gave earlier. Use the fact that

lim as x->0 sin(ax)/a = 1
and
lim as x->0 a/sin(ax)=1