Question

Help Please! Attachment to come...

Answer 1

Find the limit as x approaches 1 for both of these equations.

Answer 2

For C. I would try L'Hospital's Rule, Have you tried direct substitution on D.?

Answer 3

I'll try give L'Hospital's rule a try for C.

And direct substitution in D leads to a division by 0...

Answer 4

Direct substitution in (D) does not lead to division by 0... try again

Answer 5

Oops! you're right! I've got an answer of 1 for D... but could some one please confirm it because i'm unsure if it's right...

Answer 6

for D
sin(pi)/1 + 2 = 0 + 2
=2

Answer 7

second one is \(\pi+2\) from your eyeballs. first one, i have no idea

Answer 8

whoa it says x goes to 1??
i thought it was zero, ignore my response

Answer 9

Ok...I'm trying L'Hospital's Rule for the first one right now, as @Tonks suggested...i'll let you know the answer i get and could someone please confer it?

Answer 10

*confirm

Answer 11

you are sure it is x goes to one for the second one?

Answer 12

for the first one i'm getting sin(1)/2pi^2...i'm not sure what to do about the sin(1)?

Answer 13

OK, for C:
First check that the ratio is an indeterminate form of 0/0 or inf/inf

0/0, check

Next take the derivative of the top and the bottom:
1-cos(4pi.x) has a derivative of 4pi*sin(4pi*x)
the denominator has a derivative of 4pi^2(x-1)
When evaluated at 1, it is still 0/0, so try L'H again...

Answer 14

and @satellite73 yes i'm sure x goes to one for both of them

Answer 15

ok then second get 2 from your eyeballs. since \(\sin(\pi)=0\)

Answer 16

Poofy, have you run a second round of L'H yet?

Answer 17

first one do what @Tonks said and use l'hopital twice since you have \((x-1)^2\) in the denominator, i.e. 1 is a zero of multiplicity 2

Answer 18

just distribute denominator after the first L'Hopital's

Answer 19

@PoofyPenguin Let me know if you need help with applying L'Hopital's Rule twice

Answer 20

yes please! some help would be great! I'm very new to I'h rule and derivatives of trig functions...

Answer 21

when i try it it doesn't seem to work out for me...

Answer 22

Okay, when you get a for of zero over zero (or infinity over infinity), you can;t know what the answer will be, it is indeterminate. So instead, you look at the rate at which the parts of the ratio are approaching zero (or infinity) and that tells you which part, the numerator or the denominator will "win" by getting to zero faster (or infinity).

So when f(a)/g(a) produces 0/0 like in C., take the derivative of the top and bottom and try evaluating again. as long as you get another indeterminate form, you can keep taking the derivative. I'll show you the next step in my next post.

Answer 23

From Before:
First check that the ratio is an indeterminate form of 0/0 or inf/inf

0/0, check

Next take the derivative of the top and the bottom:
1-cos(4pi.x) has a derivative of 4pi*sin(4pi*x)
the denominator has a derivative of 4pi^2(x-1)
When evaluated at 1, it is still 0/0, so try L'H again...

L'H again:
4pi*sin(4pi*x) has derivative of 16pi^2*cos(4pi*x)
4pi^2(x-1) has derivative of 4pi^2
when evaluated at 1, you get...

Answer 24

4

Answer 25

alright! I'm a little slow, and it'll take me a little while to process but i understood everything you explained completely! Thanks so much for your help! :D

Answer 26

Your welcome

Answer 27

*you're