Question

Can anyone please help me answer this question?

Find the limit of p(x)= (x^4 - x^3 - 1)/ x^2 (x + 1) as x approaches -3

Answer 1

Step-by-step explanation:To find the limit of p(x) as x approaches -3, we can first simplify the expression by factoring the numerator:

p(x) = (x^4 - x^3 - 1) / x^2(x + 1)

= [(x - 1)(x^3 + x^2 - x - 1)] / [x^2(x + 1)]

Now, when x approaches -3, the denominator of the fraction becomes zero, which means we have an indeterminate form of the type 0/0. To evaluate the limit, we can use L'Hopital's rule, which states that if we have an indeterminate form of the type 0/0 or infinity/infinity, we can take the derivative of the numerator and denominator separately and then evaluate the limit again.

Taking the derivative of the numerator and denominator, we get:

p'(x) = [(3x^2 - 2x - 1)(x^2 + 2x) - 2(x - 1)(2x + 1)] / [x^3(x + 1)^2]

Now, plugging in x = -3 into the derivative, we get:

p'(-3) = [(3(-3)^2 - 2(-3) - 1)((-3)^2 + 2(-3)) - 2((-3) - 1)(2(-3) + 1)] / [(-3)^3((-3) + 1)^2]

= [28 - 44] / [(-3)^3(-2)^2]

= -16 / 108

= -4 / 27

Since the derivative is defined and nonzero at x = -3, we can conclude that the original limit exists and is equal to the limit of the derivative, which is:

lim x->-3 p(x) = lim x->-3 [(x - 1)(x^3 + x^2 - x - 1)] / [x^2(x + 1)]

= p'(-3)

= -4 / 27

Therefore, the limit of p(x) as x approaches -3 is equal to -4/27.

Answer 2

`\lim_(x \to -3)p(x) =-(107)/(18)`

Explanation:

Given the function
`p(x)=(x^4-x^3-1)/(x^2(x+1))`

Let's give the expressions in the numerator and denominator their own function names so they are easy to refer to:

n, for numerator:
`n(x)=x^4-x^3-1`

d, for denominator:
`d(x)=x^2(x+1)`

So
`p(x)=(n(x))/(d(x))`

Now, we want the limit of p(x) as x goes to -3.

`\lim_(x \to -3)p(x) =\lim_(x \to -3)(n(x))/(d(x))`

For limits of quotients, it is important to analyze the numerator and the denominator.

Take a moment to observe that inputting -3 into the denominator is defined and does not equal zero:
`d(-3)=(-3)^2((-3)+1)=-18\\e0`

Also, observe that inputting -3 into the numerator is defined:
`n(-3)=(-3)^4-(-3)^3-1=81+27-1=107`

Importantly, both functions n & d are polynomials, which are functions that are continuous over
`\mathbb{R}`.

Since both functions n & d are continuous, both n & d are defined at
`x=-3`, and
`d(-3)\\e0`, then the limit of the quotient is the quotient of the limits:

`\lim_(x \to -3)(n(x))/(d(x))=( \lim_(x \to -3)n(x))/( \lim_(x \to -3)d(x))`

From here, again, since n & d are continuous over
`\mathbb{R}` and defined at the limit,
`\lim_(x \to -3)n(x)}=n(-3)` and
`\lim_(x \to -3)d(x)}=d(-3)`.

Therefore,

`\lim_(x \to -3)p(x) =\lim_(x \to -3)(n(x))/(d(x))=( \lim_(x \to -3)n(x))/( \lim_(x \to -3)d(x))=(n(-3))/(d(-3))=(107)/(-18)=-(107)/(18)`