Question

lim as (x->0) of (tan(2 x))/(tan(pi x)) = 2/pi

can some please explain the process of arriving at this limit without using the l'hospital rule

Answer 1

l'hospital's rule is VERY important in calculus... and i WOULD use it for this problem...

Answer 2

we are not allowed to use it thus far in calculus because we havent been taught it. Im in calc one second week. i cant find away around it though

Answer 3

You can use the fact that tan(c*x) is approximated by the line c*x when x is close to zero

Answer 4

@zdstrong you have to first rewrite the numerator and denominator into terms of sin and cos

Answer 5

you can do it shi's way also... longer however:)

Answer 6

so \[\frac{ \tan(2x) }{\tan(\pi x) } = \frac{ \frac{ \sin(2x) }{ \cos(2x) } }{ \frac{ \sin(\pi x) }{ \cos(\pi x) } } = \frac{ \frac{ \sin(2x) }{ 1 }*\frac{ 1 }{ \cos(2x) } }{ \frac{ \sin(\pi x) }{ 1 }*\frac{ 1 }{ \cos(\pi x) } } \]

still trying to solve the problem?

Answer 7

@Shi you need to use sinc(x) ... not sure if you were going that way eventually...

Answer 8

haha I don't think im aware of your method Algebraic, though this method is probably how they want it solved

Answer 9

no.. afterward I just multiply the numerator by 2x/2x and the denominator by (pi x)/(pi x)

Answer 10

\[\sin (2*x) / \sin (\pi*x) * (2*x/2*x) / (\pi*x/\pi*x)\]

Answer 11

what is sinc(x)?? I dont think ive used it before and i want to learn haha

Answer 12

yes:) we're doing the same thing... the tangent thing is easier though... linearization of tan(c*x) = c*sec^2(c*x) for x=0

Answer 13

sinc(x) is sin(x) / x

Answer 14

and the limit is 1 as x->0

Answer 15

same thing you were doing I think...

Answer 16

ah, don't think I knew that, and not sure if that was taught at the level of this problem

Answer 17

it's the only other way to do it afaik... what were you going to do?

Answer 18

I was talking about the linearization of tan haha

ok just to clear everything up for op the last step looks like