Help with some limits

[Shoeberto]

I've been through calculus, but I was too lazy to take the AP test, so here I am in it again in college! So here's some limits I can't figure out. Hopefully ya'll can help me before 12:30 tomorrow.

lim     [sqrt(x-2) - sqrt(6-x)]/(x-3)
x->3

and

lim    ln[(x+1)^((1+2x)/x)/(x^2+2x+1)]
x->0

I know how these should work out but can't seem to get them with the usual methods. Also note to anyone in AP calc in high school: Take the AP test so you don't have to spend weeks on dumb stuff like this that you only really need to understand in theory in order to get through the class. It's painful.

[qwertysaur]

the second one is 0. When you subsutute 0 for your x's, you are left with ln(1) which is 0.

Not sure what sqrtmeans in the lirst one though, so can't help you there.

[Shoeberto]

sqrt means square root.

Also, I typed that natural log one up wrong. There's an x in the denominator of the power, which gives a divide by zero. I'll amend that right now.

Rationalization techniques in the first one don't really seem to result in the right thing, because the right terms don't cancel, so you end up with a divide by zero. So it's giving me all kinds of trouble. I'm having the same problems with the second one and getting divide by zero, 'cause even applying all kinds of crazy log properties, and simplifying it in all different ways, I end up with a zero in the denominator somewhere. Sos I gots no more ideas.

[Del Murder]

Have you ever heard of L Hopital's rule? Basically if you have a limit of a ratio you can take the derivative of the top and bottom pieces separately and it it will still result in the same limit.

For your first one you have:

derivative of top: 1/2*(x-2)^(-1/2) + 1/2*(6-x)^(-1/2)
derivative of bottom: 1

limit x->3 = 1/2*(3-2)^(-1/2) + 1/2*(6-3)^(-1/2)
= 1/2*(1) + 1/2*(1/sqrt(3))
= 1/2 + 1/(2*sqrt(3))

I'm not going to do the second one because it's your homework, and math is a bitch to type out.

[Del Murder]

Ok, I will tell you that there is a way to simplify the second one to

ln(x+1)/x

using factorization and dividing by a power and properties of logs.

[qwertysaur]

Did you try L'Hopitals rule for the first one. I got (3 - (3^.5)/6)

And for the second one sorry, Use the rules of logrithms to break it up. as I did below. (I will leave out the limit statement to make it easier to read.)

ln[(x+1)^((1+2x)/x)/(x^2+2x+1)]

ln[(x+1)^((1+2x)/x)] - ln(x^2+2x+1)

((1+2x)/x)ln(x+1) - ln(x^2+2x+1)

Then apply the limit statement to both parts of the problem sepetatly, and the second part ( the ln(x^2+2x+1)) will be solved. For the first part you will an awnser of 0/0, so apply L'Hopitals rule and you will get an awnser. I got 0 as my final awnser.

[Del Murder]

We got different answers. There is an easier way to factor the second one.

Hint: x^2+2x+1 = (x+1)^2

[qwertysaur]

I did not see that. But I redid my work and found an error, so my awnser of 0 was wrong. It came out to be 1.

And just a reminder for L'Hopitals rule, you must have a quotient of two functions, and whe applying the limit, you must end up with an indeterminate form. (such as 0/0)

[Shoeberto]

We used L'Hospital's rule last year, but technically since for this assignment "we" don't know derivatives I'm guessing that's not how much of the class will go about it. But I'll try those out whenever I've gotten up from my nap.

[rubah]

I realized just now that I typed that square root one into my calculator wrong, so I was looking at the wrong graph. It's actually asymptotic as x->3, so the limit doesn't exist. My bad!

[Doomie]

I know I should know this because I've done it all already, but logs were never my strong point. But I think I can vaguely remember a log rule that states that ln(x)^y = yln(x). I'm pretty sure you could use that here. You're also gonna need to clean things up a bit. I'm positive there's some sort of factoring or grouping you could do somewhere, but am way to tired to do it. Besides, I have my own homework to do. >=]

[NeoTifa]

Have you ever heard of L Hopital's rule? Basically if you have a limit of a ratio you can take the derivative of the top and bottom pieces separately and it it will still result in the same limit.

For your first one you have:

derivative of top: 1/2*(x-2)^(-1/2) + 1/2*(6-x)^(-1/2)
derivative of bottom: 1

limit x->3 = 1/2*(3-2)^(-1/2) + 1/2*(6-3)^(-1/2)
= 1/2*(1) + 1/2*(1/sqrt(3))
= 1/2 + 1/(2*sqrt(3))

I'm not going to do the second one because it's your homework, and math is a bitch to type out.

*sob* can you help me w/ derivatives? i totally suck! i have like a 35% in calc ;_;

[qwertysaur]

Have you ever heard of L Hopital's rule? Basically if you have a limit of a ratio you can take the derivative of the top and bottom pieces separately and it it will still result in the same limit.

For your first one you have:

derivative of top: 1/2*(x-2)^(-1/2) + 1/2*(6-x)^(-1/2)
derivative of bottom: 1

limit x->3 = 1/2*(3-2)^(-1/2) + 1/2*(6-3)^(-1/2)
= 1/2*(1) + 1/2*(1/sqrt(3))
= 1/2 + 1/(2*sqrt(3))

I'm not going to do the second one because it's your homework, and math is a bitch to type out.

*sob* can you help me w/ derivatives? i totally suck! i have like a 35% in calc ;_;

Sure, just start a new thread and people will awnser your questions.