MHB Find Limit of (x^3/2 - 27)/(x-9) with Sample Values

[ardentmed]

Hey guys, I would greatly appreciate some help for this question I'm working on at the moment:
"1.a) Use the values 9.1, 9.01, 9.001, 9.0001 to guess at the value of

lim x->9 (x^3/2 - 27)/(x-9)

1.b) Then, use appropriate factoring to find the limit of (a)"

For part a, I took sample values approaching 9, so x=9.1, .. x=9.0001 and ultimately guessed that the limit is 4.5.

As for b, I couldn't get a definitive answer, but I'm guessing that factoring works. Am I close?
Thanks in advance.

 

[Dethrone]

Here, we need to factor this:

$$\frac{x^{\frac{3}{2}}-27}{x-9}$$

One thing you should notice is that you can apply the difference of cubes:

$$\frac{(x^{\frac{1}{2}})^3-27}{x-9}$$

From here, you can directly apply the difference of cubes, in which case, you will be dealing with $\sqrt{x}$'s. However, if you want to deal with only integer exponents, you can make a substitution for x. What could that substitution be?

 

[ardentmed]

Here, we need to factor this:

$$\frac{x^{\frac{3}{2}}-27}{x-9}$$

One thing you should notice is that you can apply the difference of cubes:

$$\frac{(x^{\frac{1}{2}})^3-27}{x-9}$$

From here, you can directly apply the difference of cubes, in which case, you will be dealing with $\sqrt{x}$'s. However, if you want to deal with only integer exponents, you can make a substitution for x. What could that substitution be?

Wow, I never saw the corrolation there. So I ended up cancelling (x^.5 - 3) in the nominator and denominator and substituted x -> 9 and ultimately computed:

3+9+9 / 6 = 7/2.

Am I close? Thanks.

 

[Dethrone]

Close. Double-check your factoring.

 

[ardentmed]

Close. Double-check your factoring.

Just did and got the same answer. Shouldn't it be 9/2?

 

[Dethrone]

Check what you typed above. (Wink)

3+9+9 / 6 = 7/2.

Am I close? Thanks.

Yup, $$\frac{9}{2} $$is the right answer!

Though, it would be wise to note that your limit is of the form:

$$\displaystyle \lim_{x \to a} \frac{f(x)-f(a)}{x-a}$$
Where $f(x) = x^{3/2}$. You could have rewritten the limit as a derivative and solve it via power rule.

 

[ardentmed]

Check what you typed above. (Wink)
Yup, $$\frac{9}{2} $$is the right answer!

Though, it would be wise to note that your limit is of the form:

$$\displaystyle \lim_{x \to a} \frac{f(x)-f(a)}{x-a}$$
Where $f(x) = x^{3/2}$. You could have rewritten the limit as a derivative and solve it via power rule.

I know that it's 9/2 because of my answer from part one of the question. However, I can't get to 9/2 for some odd reason for part b. My factoring may be off, but I've already double checked it. Could you please point out the flaw in my work?

Thanks.

 

[Dethrone]

Your factoring is wrong. $$a^3-b^3=(a-b)(a^2+ab+b^2)$$, where $a=\sqrt{x}$ and $b=3$