help with a calculus limit problem?

Question

lim (x^3+27)/(x^2+5x+6)=
as x approaches -3

Answer 1

Simplify it:

(x^3) = (x+3)(x^2 -3x +9)
(x^2 + 5x + 6) = (x+3)(x+2)

new version:

lim[(x^2-3x+9)/(x+2)]

We have "renormalized" the problem by getting rid of the pole at x=-3

Since this new limit clearly exists, then the limit exists for the original expression.

"New" limit = 27/-1 = -27


Check "old" limit (meaning, we check the old equation with values of x getting closer and closer to -3 -- of course, we cannot check the old equation at x=-3 because that would force us to divide by zero)

-3.02, -26.647...
-3.01, -26.822...
-2.99, -27.182
-2.98, -27.367...

-3.002, -26.964...
-3.001, -26.982...
-2.999, -27.018...
-2.998, -27.036

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A proof could include the analysis of the neighbourhood of x=-3 (excluding x=-3) showing that for any small delta, there exists an epsilon such that for an x contained within -3 +/- epsilon, f(x) will always be within -27 +/- delta. i.e., the closer you get to x=-3, the closer the "answer" gets to -27.

Answer 2

.

lim[x->-3] { (x^3+27)/(x^2+5x+6) }

You must know the formula:
a^3 + b^3 = (a+b)*(a^2 - ab + b^2)
and I believe you will know how to split the middle of a quadratic equation to factorize it

So, lim[x->-3] { (x^3+27)/(x^2+5x+6) } can be simplified as,

= lim[x->-3] { (x+3)*(x^2 - 3x + 3^2) / (x+2)*(x+3) }

You can now cancel the common factor (x+3)

= lim[x->-3] { (x^2 - 3x + 3^2) / (x+2) }

Now, substituting the value of x as -3

= { (-3)^2 - (3)(-3) + (3)^2 / (-3+2) }

= { (9 + 9 + 9) / -1 }

= -27

.

Answer 3

By simplifying (x+3)(x^2-3x+9)/(x+3)(x+2)
= x^2-3x+9/x+2
when x approaches-3
=9+9+9/-1=-27

Answer 4

The answer is -27!

When you plug in a -3 you get 0 / 0 which is indeterminate. To solve this, you use L'Hopital's rule and differentiate the numerator and denominator. Then you have 3x^2 / (2x + 5). Plug in the -3 again and you get -27

Answer 5

If you just plug in -3 to the expression, the limit is 0/0-- undefined. So, use l'Hopital's rule, and take the derivative of the top and bottom parts of the expression-- you get(3x^2)/(2x+5)-- which isn't 0/0, but is the limit as x-> -3.

Answer 6

the limit is 0 because as x approaches -3 , (x^3+27) approaches 0 and (x^2+5x+6) also approaches 0( you simply make x=-3)

Answer 7

Both numerator and denominator have (x + 3) as a factor, which you can divide out (which is legitimate and your considering the limit, rather than the value at x = -3).

Answer 8

note that 4x - x^2 = x(4 - x) = x(2 + sqrt(x))(2 - sqrt(x)) now sub in to grant: lim (2 - sqrt(x)) / [x(2 + sqrt(x))(2 - sqrt(x)) (2 - sqrt(x)) / (2 - sqrt(x)) cancels to at least a million see you later as x =/= 4, and is a detachable discontinuity you may replace in 4 for x in what's left to locate the decrease a million / [x(2 + sqrt(x)] as x ==> 4 is a million / [4(2 + 2)] = a million/16 decrease as x ==> 4 = a million/16 once you initially substitue 4 for x, you get 0 / 0, it particularly is indeterminate, telling you that you'll be able to search for factors of 0 to cancel..

Answer 9

lim(x^3+27)/(x^2+5x+6)
x-->3
d/dx(x^3+27)/d/dx(x^2+5x+6)
3x^2/2x+5
3*(-3)^2/2*-3+5
-27