Originally Posted by
ecMathGeek I think for each of these we just need to rewrite the limits in a form that we can do:
lim t->0 of (t^(3))/[tan(2t)]^(3)
lim [t>0] [t^3/(sin(2t))^3]*(cos(2t))^3
We can rewrite this as two limits multiplied together:
lim [t>0] t^3/(sin(2t))^3 * lim [t>0] (cos(2t))^3
lim [t>0] t^3/(sin(2t))^3 * (1)
Now, I'll use the identity: sin(2t) = 2sint*cost
lim [t>0] t^3/(2sint*cost)^3
Rewritting this as the multiplication of limits again and then simplifying:
lim [t>0] t^3/(2sint)^3 * (1)
I'll factor the 1/2^3, and then rewrite the problem as:
1/8 lim [t>0] (t/sint)^3 = 1/8 ... Since lim [t>0] of t/sint = 1
lim t->0 of [sqrt(1+tan(t)) - sqrt(1+sin(t))]/(t^(3))
I can't think of what to do for this one. I'll have to think about it some more and maybe get back to you with an answer.
I encourage someone else to take a stab at this one!
lim x->0 (tan(6x))/(sin(2x))
This one we will treat similar to how I did the first. Any time I get multiplication or division of a cosine, I'm going to break it away and make it equal 1.
lim [x>0] (sin(6x)/cos(6x))/(sin(2x))
lim [x>0] (2sin(3x)cos(3x))/(2sin(x)cos(x)) <-- remove both cosines and reduce the 2/2.
lim [x>0] sin(3x)/sin(x) <-- Note: sin(3x) = sin(x)cos(2x) + sin(2x)cos(x)
lim [x>0] (sin(x)cos(2x) + sin(2x)cos(x))/sin(x) <-- separate this into an addition of limits
lim [x>0] sin(x)cos(2x)/sin(x) + lim [x>0] sin(2x)cos(x)/sin(x) <-- remove both cosines
1 + lim [x>0] 2sin(x)cos(x)/sin(x) <-- simplify
1+2 = 3