# Thread: Can someone check if i done the questions correctly: Limits

1. ## Can someone check if i done the questions correctly: Limits

Q1:

lim 2x^3 squareroot(x+6)
x->3

I substitute in the 3
=2(3)^3 squareroot(3+6)
=(54)(3)
=162
I'm not sure if 162 is the limit or not.. the number seems huge

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Q2:

lim (64-x^2)/(squareroot(8-x))
x-> negative infinity

First i found the greatest degree in the denominator
= squareroot of x

=((64/squareroot x)-(x^2/ squareroot x))/(squareroot (x(8/x-1))
=substitute in negative infinity
=-1/squareroot x
=0

Thanks

2. ## Re: Can someone check if i done the questions correctly: Limits

Q2 "substitute in negative infinity" -- Seriously? Please never do that. "negative infinity" is NOT a number. Plus, you don't know it the function is continuous. Why are you substituting at all?

Let's assume that as x increases without bound in the negative direction that it eventually decreases below -8. (Why did I bother to say that?)

For x < -8, $\frac{64-x^{2}}{\sqrt{8-x}}\;=\;-\sqrt{\frac{\left(64-x^{2}\right)^{2}}{8-x}}$

Does that do anything for our hunt for the limit?

3. ## Re: Can someone check if i done the questions correctly: Limits

I stared this for quite a while, I don't know how to continue on. You said so because the domain is x<8? Whats within sqrt must be >=0 and the denominator must not equal 0
can the this function be further simplified?

4. ## Re: Can someone check if i done the questions correctly: Limits

Notice that: $(64-x^2)=(8-x)(8+x)$ therefore:
$-\sqrt{\frac{(64-x^2)^2}{8-x}}=-\sqrt{\frac{(8-x)^2(8+x)^2}{8-x}}=-\sqrt{(8+x)^2(8-x)}$