# calculus

posted by .

lim (x^3-1)/(x^3+2x^2y+xy^2-x^2-2xy-y^2)
(x,y)->(1,0)

my question is can you approach (1,0) with y=x and does that change the the limit to

lim f(x,x)
(x,x)->(1,1)

in which case i get 3/4 as the limit. and if this is not how i go about doing it can you point me in the right direction?

• calculus -

You cannot approach along y = x because that relationship does not apply at (1,0).

Try approaching along y = 0

Lim (x^3-1)/(x^3+2x^2y+xy^2-x^2-2xy-y^2)
= Lim (x^3-1)/(x^3 -x^2)
x->1
= Lim 3x^2/(3x^2 -2x)
x->1
= 3

I used L'Hopital's rule.

• calculus -

alright thanks, my teacher never showed us that if you set one of the variables equal to zero you can make it a one variable function. so say i approach x=0 and find the limit of that to be infinity. does the original function f(x,y) not exist?

• calculus -

or does x=0 not work because x does not equal zero on the point (1,0). cause my teacher was fairly lazy in class and used the point (0,0) for every example.

• calculus -

<<or does x=0 not work because x does not equal zero on the point (1,0)? >>
It does not work because x is not 1 at y = 0, where they want the limit evaluated.

• calculus -

then what can i approach by as well cause i also tried y=x-1 and that was just a mess.

## Similar Questions

1. ### calc

need to find: lim as x -> 0 of 4(e^2x - 1) / (e^x -1) Try splitting the limit for the numerator and denominator lim lim x->0 4(e^2x-1) (4)x->0 (e^2x-1) ______________ = ________________ lim lim x->0 e^X-1 x->0 e^x-1 …
2. ### Calculus

Find the indicated limits. If the limit does not exist, so state, or use the symbol + ∞ or - ∞. f(x) = { 2 - x if x ≤ 3 { -1 + 3x - x^2 if x > 3 a) lim 3+ f(x) x->3 b) lim 3- f(x) x->3 c) lim f(x) x->3 …
3. ### Calculus

Find the indicated limits. If the limit does not exist, so state, or use the symbol + ∞ or - ∞. f(x) = { 2 - x if x ≤ 3 { -1 + 3x - x^2 if x > 3 a) lim 3+ f(x) x->3 b) lim 3- f(x) x->3 c) lim f(x) x->3 …
4. ### Calculus

lim (x^3-1)/(x^3+2x^2y+xy^2-x^2-2xy-y^2) (x,y)->(1,0) alright i approached this at y=0. and got the limit as 3. where else can i approach this from to see if the limit really exist which im hoping it doesn't. i tried y=x-1 and that …
5. ### Calculus

I have two similar problems that I need help completing. Please show all your work. Question: Find the limit L. Then use the å-ä definition to prove the limit is L. 1. lim (2x+5) x->3 2. lim 3 x->6 Thank you for your anticipated …
6. ### Calculus. Limits. Check my answers, please! :)

4. lim (tanx)= x->pi/3 -(sqrt3) 1 (sqrt3) ***-1 The limit does not exist. 5. lim |x|= x->-2 -2 ***2 0 -1 The limit does not exist. 6. lim [[x]]= x->9/2 (Remember that [[x]] represents the greatest integer function of x.) 4 …
7. ### Check my CALCULUS work, please! :)

Question 1. lim h->0(sqrt 49+h-7)/h = 14 1/14*** 0 7 -1/7 Question 2. lim x->infinity(12+x-3x^2)/(x^2-4)= -3*** -2 0 2 3 Question 3. lim x->infinity (5x^3+x^7)/(e^x)= infinity*** 0 -1 3 Question 4. Given that: x 6.8 6.9 6.99 …
8. ### Calculus

Find the limit. lim 5-x/(x^2-25) x-->5 Here is the work I have so far: lim 5-x/(x^2-25) = lim 5-x/(x-5)(x+5) x-->5 x-->5 lim (1/x+5) = lim 1/10 x-->5 x-->5 I just wanted to double check with someone and see if the answer …
9. ### Calculus (lim)

consider the statement lim x³-6x²+11x-6 / x-1 = 2 x->1 Using the definition of the limit, state what must be true for the above limit to hold, that is, for every ..., there is ..., so that.... Use a specific function and limit …
10. ### Pre-Calculus

Which of the following shows the correct notation for “The limit of x^2 - 1 as x approaches 3. A. lim x^2-1 x->3 B. lim3 x->x^2-1 C. lim(x^2-3) x->x^2-1 D. lim(x^2-1) x->3 Thank you

More Similar Questions