Evaluating Limit: $\sqrt[3]{x + 2} \div (x + 8)$ as $x \rightarrow -8$

[jmed]

Homework Statement

evaluate limit as x approaches -8 of cubed root of x + 2 divided by x +8

Homework Equations

The Attempt at a Solution

 

[jmed]

I have this so far... (1/3x^-2/3 * x + 8) - (x^1/3 + 2 * 1) all over (x + 8)^2

 

[Mark44]

Is this your limit?
\lim_{x \to -8}\frac{\sqrt[3]{x} + 2}{x + 8}

Or is it this?
\lim_{x \to -8}\frac{\sqrt[3]{x + 2}}{x + 8}

As a tip, you can see my LaTeX script by double-clicking either of the limit expressions above.

What you wrote is ambiguous since it does not make clear what's in the cube root. From your work, it appears that the first limit expression above is the one you are trying to evaluate.

The work you did is completely wrong. You apparently are attempting to use L'Hopital's Rule, but are mistakenly using the quotient rule in doing so. This is not at all how L'Hopital's Rule works or should be used.

Another technique that is simpler, IMO, is to factor the denominator, x + 8, into
(x^{1/3} + 2)(x^{2/3} - 2x^{1/3} + 4)

After getting rid of common factors in the numerator and denominator you should be able to evaluate the limit pretty easily.

 

[jmed]

well we have not learned the L'Hopitals Rule yet...we have only learned up to product and quotient rules. and it is the first equation I am evaluating.

 

[Mark44]

So why are you using either the product rule or quotient rule to evaluate a limit?

 

[jmed]

that is what we just learned...

 

[Mark44]

I'm sure that's not what you were taught -- that to take a limit you take the derivative.

 

[jmed]

I guess I am just confused...would you multiply by the conjugate of the numerator?

 

[Mark44]

No. Look at my response back in post #2.

 

[jmed]

i did...not sure how you got that..

 

[Mark44]

I factored x + 8, treating it as the sum of two cubes - (x^1/3)³ + 2³.

The general formula is a³ + b³ = (a + b)(a² - ab + b²).

There's also a formula for the difference of two cubes - a³ - b³ = (a - b)(a² + ab + b²).

 

[jmed]

ahhh, ok. now would I plug in -8 for x?

 

[Mark44]

Factor the denominator as described in post #3. That should enable you to do some simplification. Then take the limit.

 

[jmed]

awesome! thanks so much!

 

[Mark44]

Sure - you're welcome!

 

[jmed]

To be sure... the answer is as x approaches -8 the limit is 1/4? Just want to check my answer.

 

[Mark44]

To be sure... the answer is as x approaches -8 the limit is 1/4? Just want to check my answer.

I get 1/12.