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find the limit as it approaches -3 in the equation (6x+9)/x^4+6x^3+9x^2


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The limit of (6x+9)/x^4+6x^3+9x^2 as x approaches -3 does not exist.  If you factor the rational expression you get:

 (6x+9)/[x^4+6x^3+9x^2]  =   3(2x+3) / [ (x^2)(x+3)(x+3) ]

If we let x go to -3 (from both sides) the numerator is well behaved and approaches -9, however the denominator goes to zero (but is always positive, approaching from either the left or the right).  So the entire rational expression will approach negative infinity as x goes to -3.  Since there is no specific value for the limit, we say that the limit does not exist.

(6X+9) / (X4 + 6X3 +9X)=
6 ( X +3 )/ X2 ( X2 +6X +9)
6 ( X+3)/ X2 ( X +3 )2 =
lim x = -3 6 / X2 ( X + 3) =limx= -3 6/ (9 ( X + 3)= 2/(3 *0) = undefined

Let's re-write the expression first :

Lim (x-->-3) [3(2x+3)/x2 (x+3)2]

By looking at the term (x+3)2, that term approaches zero... Which would make the denominator zero... Meaning the function will approach infinity. The demonimator will always be positive. The numerator will have a negative value as you approach from the left (negative infinity)... but a positive value as you approach from the right (positive infinity). Since the limits from the left and right do not equal, the limit for the expression DOES NOT EXIST.


"...but a positive value as you approach from the right (positive infinity)."
This is incorrect.  The numerator ( 3(2x+3) ) is negative as x approaches -3 from either side, so the overall expression approaches negative infinity for both the left and right limit.  A limit that approaches infinity is not defined and therefore the limit does not exist.

limit as it approaches -3 in the equation (6x+9)/x^4+6x^3+9x^2

We need to see if the numerator and denominator are both 0, +-infinity or either is infinity or if the denominator is zero.  Cases 1 and 2 (both zero or bothe infinite) can be checked further using L'hopitals rule, namely if lim x-->a f(a)/g(a) = 0/0 or inf/inf, we can take the derivative of both f(x) and g(x) and then reevaluate namely lim x ---->a = f'(a)/g'(a) = lim x----> a f(x)/g(x)

So for our problem limit:  lim x ---> -3 (6x+9)/x^4+6x^3+9x^2

The numerator is 6(-3) + 9 = -18+9 = -9

The denominator is -3^4 + 6*-3^3 + 9*-3^2 = 81 - 162 + 81 = 0.

Thus our limit is lim x-->-3  -9/0 = negative infinity.

No other calculation is needed!! 




l'Hôpital's rule also called Bernoulli's rule, uses derivatives to help evaluate limits involving indeterminate forms: and you can onely use it if lim f =lim g = 0 or +–8 and other conditions. However, this condition is not satisfied in this situation so you cannot use this rule. Sorry

I take it you already tried plugging 3 into the function to see if the function exists there, right?

When you try to solve this, think about what the question is asking. You want to find the expression "in the limit that" x=-3. Have you tried plugging in numbers that are very close to -3 and solving the expression? What do you get if x=-3.5, or -2.5?

If you have a graphing calculator, try plotting the expression y=(6x+9)/(x4+6x3+9x2). The limit might be obvious if you represent the expression graphically.