# Help with limits to infinity

• Jun 11th 2011, 08:03 PM
ftmichael
Help with limits to infinity
I'm in a beginning calculus class after not touching math of any sort for 4 years. I've forgotten a lot of the more complicated algebra, and I'm struggling with simplifying limits right now. I'm doing my homework right now and coming across a lot of problems I'm getting wrong and I can't understand why, so I am asking for help.

first problem:

1.) $\displaystyle lim x\to \infty \frac {\sqrt{9x^6 - x}}{x^3 + 1}$

To work this one out, I multiplied the top and bottom by $\displaystyle \sqrt{9x^6 - x}$,
so I get
$\displaystyle \frac{9x^6-x}{(x^3+1)(\sqrt{9x^6-x})}$

From there, I divided everything by $\displaystyle x^3$ since that's the highest x value in the denominator in order to get rid of some of the numbers. I wind up with:
$\displaystyle \frac{9x^3-0}{ (1+0)(\sqrt{9-0} }$
Which is $\displaystyle \frac{9x^3}{3 }$ which equals
$\displaystyle 3x^3$

The back of my book however says this answer should just be 3. Why is the x^3 cancelled?
---------------------

Here's another I got wrong:
$\displaystyle lim x\to \infty \sqrt {9x^2 + x} - 3x$

All I could think to do was to divide everything by X but I don't even think you're allowed to do that without a denominator (are you?). Anyway when I do that I get
$\displaystyle \sqrt{9+0} - 3 = 0$

Apparently the answer is supposed to be 1/6, how do you get this?
• Jun 11th 2011, 08:06 PM
dwsmith
$\displaystyle \lim_{x\to\infty}\frac{\sqrt{9x^6}}{x^3}=3$

When looking at a fraction of a limit to infinity, you need to only look at the highest power terms of the denominator and numerator.
• Jun 11th 2011, 08:12 PM
dwsmith
For the second one, multiply and divide by the conjugate.
• Jun 11th 2011, 08:25 PM
Prove It
Quote:

Originally Posted by ftmichael
I'm in a beginning calculus class after not touching math of any sort for 4 years. I've forgotten a lot of the more complicated algebra, and I'm struggling with simplifying limits right now. I'm doing my homework right now and coming across a lot of problems I'm getting wrong and I can't understand why, so I am asking for help.

first problem:

1.) $\displaystyle lim x\to \infty \frac {\sqrt{9x^6 - x}}{x^3 + 1}$

To work this one out, I multiplied the top and bottom by $\displaystyle \sqrt{9x^6 - x}$,
so I get
$\displaystyle \frac{9x^6-x}{(x^3+1)(\sqrt{9x^6-x})}$

From there, I divided everything by $\displaystyle x^3$ since that's the highest x value in the denominator in order to get rid of some of the numbers. I wind up with:
$\displaystyle \frac{9x^3-0}{ (1+0)(\sqrt{9-0} }$
Which is $\displaystyle \frac{9x^3}{3 }$ which equals
$\displaystyle 3x^3$

The back of my book however says this answer should just be 3. Why is the x^3 cancelled?

L'Hospital's Rule works nicely in both cases. If you have an indeterminate form $\displaystyle \displaystyle \frac{0}{0}$ or $\displaystyle \displaystyle \frac{\infty}{\infty}$, then

$\displaystyle \displaystyle \lim_{x \to c}\frac{f(x)}{g(x)} = \lim_{x \to c}\frac{f'(x)}{g'(x)}$

The first is already in $\displaystyle \displaystyle \frac{\infty}{\infty}$ form, and the second can be made into that form using this transformation...

\displaystyle \displaystyle \begin{align*} \sqrt{9x^2 + x} - 3x &= \ln{\left(e^{\sqrt{9x^2 + x} - 3x}\right)} \\ &= \ln{\left(\frac{e^{\sqrt{9x^2 + x}}}{e^{3x}}\right)}\end{align*}

Recall that because of the continuity of the logarithm, $\displaystyle \displaystyle \lim_{x \to c}\ln{[f(x)]} = \ln{\left[\lim_{x \to c}f(x)\right]}$, and the "stuff" inside the logarithm is now of the form $\displaystyle \displaystyle \frac{\infty}{\infty}$. So you can use L'Hospital's Rule there too :)
• Jun 11th 2011, 08:47 PM
ftmichael
Quote:

Originally Posted by Prove It
L'Hospital's Rule works nicely in both cases. If you have an indeterminate form $\displaystyle \displaystyle \frac{0}{0}$ or $\displaystyle \displaystyle \frac{\infty}{\infty}$, then

$\displaystyle \displaystyle \lim_{x \to c}\frac{f(x)}{g(x)} = \lim_{x \to c}\frac{f'(x)}{g'(x)}$

The first is already in $\displaystyle \displaystyle \frac{\infty}{\infty}$ form, and the second can be made into that form using this transformation...

\displaystyle \displaystyle \begin{align*} \sqrt{9x^2 + x} - 3x &= \ln{\left(e^{\sqrt{9x^2 + x} - 3x}\right)} \\ &= \ln{\left(\frac{e^{\sqrt{9x^2 + x}}}{e^{3x}}\right)}\end{align*}

Recall that because of the continuity of the logarithm, $\displaystyle \displaystyle \lim_{x \to c}\ln{[f(x)]} = \ln{\left[\lim_{x \to c}f(x)\right]}$, and the "stuff" inside the logarithm is now of the form $\displaystyle \displaystyle \frac{\infty}{\infty}$. So you can use L'Hospital's Rule there too :)

Thanks for all that help, but I haven't learned this rule yet and I don't feel comfortable enough trying stuff outside of what I've already been taught. Thanks for all that effort though!

DWsmith:

Thanks for the first reply, I realized i made a stupid mistake on the first problem i posted and after working that out I got 3. Conjugating first was all I needed for the second problem as well.

Another question, How do you find horizontal and vertical asymptotes? I know for horizontal you have to make the limit of a function + or - infinity, and I'd assume for vertical it's the limit of a function when x approaches 0? I can provide a practice problem if it would be easier to grasp what im saying.
• Jun 11th 2011, 09:59 PM
dwsmith
Quote:

Originally Posted by ftmichael
Thanks for all that help, but I haven't learned this rule yet and I don't feel comfortable enough trying stuff outside of what I've already been taught. Thanks for all that effort though!

DWsmith:

Thanks for the first reply, I realized i made a stupid mistake on the first problem i posted and after working that out I got 3. Conjugating first was all I needed for the second problem as well.

Another question, How do you find horizontal and vertical asymptotes? I know for horizontal you have to make the limit of a function + or - infinity, and I'd assume for vertical it's the limit of a function when x approaches 0? I can provide a practice problem if it would be easier to grasp what im saying.

Vertical asymptote occur when the denominator is zero.
Horizontal--you need to check the limits at negative and positive infinity.
Slant--when the highest degree of the numerator is greater than the highest degree of the denominator. To solve, you do long division and disregard the remainder.
• Jun 11th 2011, 10:11 PM
Also sprach Zarathustra
Quote:

Originally Posted by ftmichael
I'm in a beginning calculus class after not touching math of any sort for 4 years. I've forgotten a lot of the more complicated algebra, and I'm struggling with simplifying limits right now. I'm doing my homework right now and coming across a lot of problems I'm getting wrong and I can't understand why, so I am asking for help.

first problem:

1.) $\displaystyle lim x\to \infty \frac {\sqrt{9x^6 - x}}{x^3 + 1}$

To work this one out, I multiplied the top and bottom by $\displaystyle \sqrt{9x^6 - x}$,
so I get
$\displaystyle \frac{9x^6-x}{(x^3+1)(\sqrt{9x^6-x})}$

From there, I divided everything by $\displaystyle x^3$ since that's the highest x value in the denominator in order to get rid of some of the numbers. I wind up with:
$\displaystyle \frac{9x^3-0}{ (1+0)(\sqrt{9-0} }$
Which is $\displaystyle \frac{9x^3}{3 }$ which equals
$\displaystyle 3x^3$

The back of my book however says this answer should just be 3. Why is the x^3 cancelled?
---------------------

Here's another I got wrong:
$\displaystyle lim x\to \infty \sqrt {9x^2 + x} - 3x$

All I could think to do was to divide everything by X but I don't even think you're allowed to do that without a denominator (are you?). Anyway when I do that I get
$\displaystyle \sqrt{9+0} - 3 = 0$

Apparently the answer is supposed to be 1/6, how do you get this?

$\displaystyle \lim_{x\to\infty}\frac{\sqrt{9x^6-x}}{x^3+1}$

I will prove by definition that the above limit equals to 3.

$\displaystyle (\forall \varepsilon >0)(\exists\Delta \in\mathbb{R})(\forall x>\Delta ):|f(x)-L|<\varepsilon$

Where $\displaystyle f(x)=\frac{\sqrt{9x^6-x}}{x^3+1}$ and $\displaystyle L=3$.

Proof:

Let be $\displaystyle \varepsilon >0$ given.

$\displaystyle |\frac{\sqrt{9x^6-x}}{x^3+1}-1|=|\frac{\sqrt{9x^6-x}-3x^3-3}{x^3+1}|\overset{x\geq \sqrt[5]{\frac{1}{9}}}{<}{|\frac{3x^3-3x^3-3}{x^3+1}|}<|\frac{3}{x^3}|<\varepsilon$

$\displaystyle |\frac{3}{x^3}|<\varepsilon$

or:

$\displaystyle 3<x^3\varepsilon$

$\displaystyle \frac{3}{\varepsilon }<x^3$

or:

$\displaystyle x>\sqrt[3]{\frac{3}{\varepsilon }}$

We set $\displaystyle \sqrt[3]{\frac{3}{\varepsilon }}$ be our delta.

$\displaystyle \Delta =\sqrt[3]{\frac{3}{\varepsilon }}$

Hence,

$\displaystyle (\forall \varepsilon >0)(\exists \sqrt[3]{\frac{3}{\varepsilon }} \in\mathbb{R})(\forall x>\sqrt[3]{\frac{3}{\varepsilon }}):|f(x)-3|<\varepsilon$
• Jun 12th 2011, 12:34 AM
Also sprach Zarathustra
Quote:

Originally Posted by dwsmith
For the second one, multiply and divide by the conjugate.

Lets do it!

$\displaystyle \lim_{x\to\infty}\sqrt{9x^2+x}-3x=?$

$\displaystyle \sqrt{9x^2+x}-3x=\frac{(\sqrt{9x^2+x}-3x)(\sqrt{9x^2+x}+3x)}{\sqrt{9x^2+x}+3x}=\frac{9x^ 2+x-9x^2}{\sqrt{9x^2+x}+3x}=$

$\displaystyle \frac{\frac{x}{x}}{\frac{\sqrt{9x^2+x}+3x}{x}}=\fr ac{1}{\sqrt{\frac{9x^2+x}{x^2}}+\frac{3x}{x}}=\fra c{1}{\sqrt{9+\frac{1}{x}}+3}$

THERE IS STRANGE LATEX ERROR THAT I CAN'T FIND!

Now,

$\displaystyle \lim_{x\to\infty}\sqrt{9x^2+x}-3x=\lim_{x\to\infty}\frac{1}{\sqrt{9+\frac{1}{x}}+ 3}=\frac{1}{\sqrt{9+0}+3}=\frac{1}{3+3}=\frac{1}{6 }$