limits questions/.?

Question

lim y3 + 8 / y + 2
y>2


lim square root of x - 1 / x - 1
x > 1



>

can u solve it to those who know well in math especially calculus... the answer should not be 0/0.. please help me..

Answer 1

Frst, when you ask questions, use parentheses to avoid confusion in the meaning of your equations.

I suppose you mean lim ( x -> - 2) (y3 + 8) /( y + 2) y approaching -2, not 2.

If we do a long division, we see that (y^3 + 8) = (y +2) (y^2 -2y + 4) . Therefore, for every y different from 2 we have

(y3 + 8) /( y + 2) = y^2 -2y + 4. Now, we have no indetermination, all we have to do is plug y =-2 in the right han side. We get lim x -> - 2) (y3 + 8) /( y + 2) = 12

As for the second, I guess you mean lim ( x -> 1) sqrt(x -1)/(x-1). Observe that x -1 = (sqrt(x) - 1)(sqrt(x) + 1). Therefore, for every x differebt from 1 we have

sqrt(x -1)/(x-1) = 1/(sqrt(x) + 1). We don't have any indetermination any more, and all we have to do is plug in x =1 We get lim ( x -> 1) sqrt(x -1)/(x-1) = 1/2

Answer 2

Should the first one be (y^3 - 8)/(y-2)? If not you just plug 2 in for y and you'll get (2^3+8)/(2+2) = (8+8)/4 = 16/4 = 4

If it's (y^3 - 8)/(y-2) you factor y^3 -8:
y^3 - 8 = (y-2)(y^2 + 2y + 4)
so now you have lim(y-2)(y^2 + 2y + 4)/(y-2)
we have y-2 on the top and bottom so they cancel each other out, and we have
lim(y^2 + 2y + 4)/1
y->2
and now we can plug 2 in for y and get
2^2+2*2 + 4 = 4 + 4+ 4 = 12

Do the same thing for the second except you don't have to factor, x-1 is already on top and bottom so essentially have:
lim sqrt(1)
x->1
so the answer is 1


L'Hopital's Rule does not work for the first one. It must be of the form 0/0 or infin/infin (or infin*0). And I'm assuming you havn't learned that yet.

Answer 3

in the first one, we can just plug in 2 because this won't result in division by zero (or anything else strange, like sqrt of a negative number), so we get:

lim (2)^3 + 8/2 + 2
y>2

lim 8 + 4 + 2
y>2

lim 14
y>2

now, for the second one we can't just plug in 1 because this will result in 0/0 which is undefined, but we can cancel the numerator and denominator, because, remember, x is never actually 1, it's only getting very, very, very close to 1. Because of this we won't actually divide by zero, so we can cancel, so we get:

lim √[(x-1)/(x-1)]
x>1

so just cancel the numerator and denominator and you get:

lim √1
x>1

lim 1
x>1

and the limit of a constant is just the constant, so our final answer is: 1

This page might help you understand limits better:

http://tutorial.math.lamar.edu/AllBrowsers/2413/ComputingLimits.asp

Answer 4

1) (Y^3 + 8)/(y + 2), as y --> 2, this = 16/4 = 4

2) [(x - 1)^1/2]/(x - 1) --> +infinity, after differentiating numerator and denomenator (L'hopital's Rule)

Answer 5

Using L'Hospital's Rule, you take the first derivative of the numerator and the first derivative of the denominator, and there is your answer:

3y², as y -> 2, the answer is 12.

I leave the second one for you to work out. :-)

Answer 6

Actually, L'Hopital's Rule isn't even necessary here. Just do the division to simplify, and you don't even need to worry about the indeterminate forms.

Answer 7

in the first question, the answer is 4