What is the limit as x approaches 0 for the function (x^3-x)/(x^3-4x)?

Answer 1

Since it has the form 0/0 at x=0, use L'Hopital's rule and find derivatives of both expressions, getting

(3x²-1)/(3x²-4)

which is 1/4 when x=0, so the limit is 1/4. graph it to verify.

Answer 2

>1

Answer 3

1

Answer 4

Factor out an x from both numerator and denominator to get
(x² - 1)/(x² - 4) So at x = 0 this becomes 1/4.
Notice that the original function does --not-- become 1/4 at x=0, it only approaches it as a limit. At x = 0 the original function is undefined.

Doug

Answer 5

1/4

Answer 6

(x^3 - x) / (x^3 - 4x)
= x(x^2 - 1) / x(x^2 - 4)
= (x^2 - 1) / (x^2 - 4)
as x approaches 0 you will have -1 / -4 or 1/4.

Answer 7

Numerator ======== x^2(x-a million) + 4(x-a million) (x^2 + 4)*(x-a million) be conscious that the two numerator and denominator contain x - a million. whilst x = a million, you get 0/0 which you would be able to no longer do. for this reason you ought to use L'wellness facility's Rule. Differentiate numerator 3x^2 - 2x +4 Differentiate the denominator a million Now take your shrink. 3(a million)^2 - 2(a million) + 4 = 5 Your answer must be d.

Answer 8

(x^3-x)/(x^3-4x)
<==>
x(x^2 - 1)/x(x^2 - 4)
(x - 1)(x + 1)/(x -2)(x + 2)
as x-->0 you have
-1(1)/(-2)(+2)
-1/-4
1/4