Limit with Square Root

[marirak]

Hi all!

Total newb here, and I was wondering whether someone could help me with this limits problem. The correct answer is 3, but I can't get this answer.

lim(x->infinity) ((9x^6 - x)^(1/2))/(x^3 + 1)

So far I've multiplied top and bottom by ((1/x^9)^(1/2))/(1/x^3) so that it's still eq to 1 but I can get in under that radical in the numerator.

I then get (((9/x^3)-(1/x^8))^(1/2))/(1+(1/x^3))

Thus, when I substitute infinity for x, I get ((0-0)^(1/2))/(1+0), so as x approaches infinity, the limit is 0. Which is definitely not 3. :/

I can see how (9^(1/2))/1 would get me the 3, so I know something is wrong with the way I'm reducing those x's but I'm doing everything my professor showed in class and can't figure out how to get that 9 on its own.

Thanks so much!!

 

[lookagain]

Hi all!

Total newb here, and I was wondering whether someone could help me with this limits problem.
The correct answer is 3, but I can't get this answer.

lim(x->infinity) ((9x^6 - x)^(1/2))/(x^3 + 1)

So far I've multiplied top and bottom by ((1/x^9)^(1/2))/(1/x^3) $\displaystyle \ \ \ \ \$<------ This is not what you mean. $\displaystyle \ \$ See below.*

so that it's still eq to 1 but I can get in under that radical in the numerator.

I then get (((9/x^3)-(1/x^8))^(1/2))/(1+(1/x^3))

Thus, when I substitute infinity for x, I get ((0-0)^(1/2))/(1+0), so as x approaches infinity, the limit is 0.
Which is definitely not 3. :/

I can see how (9^(1/2))/1 would get me the 3, so I know something is wrong with the way I'm reducing
those x's but I'm doing everything my professor showed in class and can't figure out how to get that 9 on its own.

* You are to multiply the numerator by [1/(x^6)]^(1/2), *not* [1/(x^9)]^(1/2),
and you are to multiply the denominator by 1/(x^3).

 

[marirak]

Thank you! I was wondering about that, but then you're not multiplying by 1, and so aren't you changing the function?

EDIT: Ohhhh, wow. Square root of x^6 is x^3 because exponents are additive...I was thinking sq root of 9 = 3. Silly

 

[HallsofIvy]

Since x is going to infinity the highest powers will dominate. So you can ignore other powers:
Instead of looking at all of $\displaystyle \frac{(9x^6- x)^{1/2}}{x^3+1}$, look at $\displaystyle \frac{(9x^6)^{1/2}}{x^3}= \frac{3x^3}{x^3}= 3$.

If you want to be a little more rigorous, divide both numerator and denominator by $\displaystyle x^3$:
$\displaystyle \frac{\frac{(9x^6- x)^{1/2}}{x^3}}{\frac{x^3+1}{x^3}}= \frac{(\frac{9x^6}{x^6}-\frac{1}{x^6})^{1/2}}{\frac{x^3}{x^3}+ \frac{1}{x^3}}= \frac{(9- \frac{1}{x^5})^{1/2}}{1+ \frac{1}{x^6}}$

Now, as x goes to infinity, all fractions with x in the denominator will go to 0 leaving $\displaystyle \frac{(9)^{1/2}}{1}= 3$.