# Thread: Help with limits to infinity

1. ## 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?

2. $\displaystyle \lim_{x\to\infty}\frac{\sqrt{9x^6}}{x^3}=3$

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.

3. For the second one, multiply and divide by the conjugate.

4. 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

5. 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.

6. 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.

7. 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,

$\displaystyle (\forall \varepsilon >0)(\exists \sqrt[3]{\frac{3}{\varepsilon }} \in\mathbb{R})(\forall x>\sqrt[3]{\frac{3}{\varepsilon }}):|f(x)-3|<\varepsilon$

8. 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,

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