Thread: Limits

Can't remember how to do this (without L'Hopitals)

lim t->0 of (t^(3))/[tan(2t)]^(3)

lim t->0 of [sqrt(1+tan(t)) - sqrt(1+sin(t))]/(t^(3))

lim x->0 (tan(6x))/(sin(2x))

thanks for the help

Originally Posted by drain

Can't remember how to do this (without L'Hopitals)

lim t->0 of (t^(3))/[tan(2t)]^(3)

Heir.

Originally Posted by drain

Can't remember how to do this (without L'Hopitals)

lim t->0 of (t^(3))/[tan(2t)]^(3)

lim t->0 of [sqrt(1+tan(t)) - sqrt(1+sin(t))]/(t^(3))

lim x->0 (tan(6x))/(sin(2x))

thanks for the help

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

Thanks you guys. Greatly appreciated

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

This is really hard to follow at first glance. I need to work on organizing my examples better

Originally Posted by drain

Can't remember how to do this (without L'Hopitals)

lim t->0 of (t^(3))/[tan(2t)]^(3)

lim t->0 of [sqrt(1+tan(t)) - sqrt(1+sin(t))]/(t^(3))

lim x->0 (tan(6x))/(sin(2x))

thanks for the help

if latex was working i'd write up the 2nd one. It's not so I wont.

i'll will get you started though


multiply [sqrt(1+tan(t)) - sqrt(1+sin(t))]/(t^(3)) by

[sqrt(1+tan(t)) + sqrt(1+sin(t))]/[sqrt(1+tan(t)) + sqrt(1+sin(t))]]


expand the numerator and then (after a little more work simplifying) try and find the limit.