Evaluate limit using trigonometric identities.

[philippe311]

Homework Statement

use lim as x__>0 (sin(x))/(x) to evaluate the following : lim (x^(2) cot(x) + sin(x/2) - cos(x) + 1)/(x) as x__>0

The Attempt at a Solution

1- lim as x__>0 (x^(2) cot(x))/ (x)= x cotx = x (cosx)/(sinx)= (x/sinx) cosx = 1.1=1
And I think I can use L'Hospital's Rule to evaluate the previous one, right?
2- lim as x__>0 (sin(x/2))/(x)= 1/2 (sin(x/2))/(x/2)= 1/2.
3. this is where it gets hard. I tried to use L'Hospital's Rule, but I just want to see if that makes sense or not. lim as x__>0 (-cosx +1)/(x)= 0/0. using L'H, I will get lim as x__>0 sinx= 0 . so my final answer will be 1+1/2+-0= 1.5. Is this right?

 

[Mark44]

Homework Statement

use lim as x__>0 (sin(x))/(x) to evaluate the following : lim (x^(2) cot(x) + sin(x/2) - cos(x) + 1)/(x) as x__>0

The Attempt at a Solution

1- lim as x__>0 (x^(2) cot(x))/ (x)= x cotx = x (cosx)/(sinx)= (x/sinx) cosx = 1.1=1
And I think I can use L'Hospital's Rule to evaluate the previous one, right?
2- lim as x__>0 (sin(x/2))/(x)= 1/2 (sin(x/2))/(x/2)= 1/2.
3. this is where it gets hard. I tried to use L'Hospital's Rule, but I just want to see if that makes sense or not. lim as x__>0 (-cosx +1)/(x)= 0/0. using L'H, I will get lim as x__>0 sinx= 0 . so my final answer will be 1+1/2+-0= 1.5. Is this right?

Right answer, but you need to carry the limit symbol all the way until you actually take the limit. For example, in 1, you should say lim (x^(2) cot(x))/ (x)= lim x cotx = lim x (cosx)/(sinx)= lim (x/sinx) * lim cosx = 1 * 1 = 1. (All limits are as x --> 0.)

Many instructors will ding you if you write something such as (x/sinx) cosx = 1.1=1, which isn't generally true.

 

[philippe311]

Thanks for pointing out the limit symbol.
And again thanks for checking my answer.

 

[Cyosis]

It strikes me as odd that you were allowed to use l'hospital's since the exercise states 'use lim as x__>0 (sin(x))/(x) to evaluate the following'. If you could use l'hospital's why even bother with giving you that limit?

 

[tiny-tim]

Hi philippe311!

(write "->0" … it's easier to read … and try using the X² and X₂ tags just above the Reply box )

1- lim as x__>0 (x^(2) cot(x))/ (x)= x cotx = x (cosx)/(sinx)= (x/sinx) cosx = 1.1=1
And I think I can use L'Hospital's Rule to evaluate the previous one, right?
2- lim as x__>0 (sin(x/2))/(x)= 1/2 (sin(x/2))/(x/2)= 1/2.
3. this is where it gets hard. I tried to use L'Hospital's Rule, but I just want to see if that makes sense or not. lim as x__>0 (-cosx +1)/(x)= 0/0. using L'H, I will get lim as x__>0 sinx= 0 . so my final answer will be 1+1/2+-0= 1.5. Is this right?

3. is a bit difficult to read … if you're saying lim_x->0 (1-cosx)/x = lim_x->0 sinx/1 = 0, then 1 2 and 3 are all correct.

However, the question doesn't ask you to use l'Hôpital's rule except specifically for sinx/x … perhaps you should use one of the standard trigonometric identities … 1 - cosx = 2sin²(x/2)) ?