# Thread: limit of trig function

1. ## limit of trig function

lim t->0 tan6t/sin2t

I did

lim t->0 tan6t/sin2t=

lim t->0 (6sin6t/6t * 1/cos6t * 2t/2sin2t)=

6 * lim t->0 (sin6t/6t ) * lim t->0 (1/cos6t) * 1/2 * lim t->0 (2t/sin2t)=

6 * 1 * lim t->0 (1/cos6t) * 1/2 * 1

So I stick with the middle term. In the solution they have
lim t->0 (1/cos6t) = 1/1
but I do not know how they get that.

I have the formulas:

lim x->0 sinx/x = 1
lim x->0 (cosx-1)/x = 0

d/dx sinx = cosx
d/dx cosx = -sinx

Can someone help me with that?
Thanks

2. Is it possible that I can use direct substitution here, since cos6t cannot get zero?
cos(0)=1

so
lim t->0 (1/cos6t) = 1/ cos(0) = 1/1 = 1

Then
lim t->0 tan6t/sin2t = 6 * 1 * 1 * 1/2 * 1 = 3

I that right?
Thanks

3. Originally Posted by DBA
Is it possible that I can use direct substitution here, since cos6t cannot get zero?
cos(0)=1

so
lim t->0 (1/cos6t) = 1/ cos(0) = 1/1 = 1 Mr F says: Yes.

Then
lim t->0 tan6t/sin2t = 6 * 1 * 1 * 1/2 * 1 = 3

I that right?
Thanks
Yes.