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

Question

lim_-3=(6x+9)/(x⁴+6x³+9x²)

Answer 1

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.

Answer 2

If you graph the function you will see this is true.

But, since it does not approach any VALUE then the limit does not exist.

Answer 3

(6X+9) / (X⁴ + 6X³ +9X² )=

 

6 ( X +3 )/ X² ( X² +6X +9)

 

6 ( X+3)/ X² ( X +3 )^{2 =}

 

lim _{x = -3} 6 / X² ( X + 3) =lim_{x= -3} 6/ (9 ( X + 3)= 2/(3 *0) = undefined

Answer 4

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.

Answer 5

Use L'Hopital's rule

take derivatives of the numerator and denominator, at x=-3. that gives 0/0 which is undefined

so take the 2nd derivatives of numerator and denominator, at x=-3. that gives 0/48 =0

the limit as x approaches -3 is 0

Answer 6

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!! 

 

 

Answer 7

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.

Answer 8

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