Limit to Infinity of Quotient with Square Root - Reducing the Equation

[njh]

Homework Statement

Find the limit as x tends to negative infinity of the quotient.

Relevant Equations

$\lim_{x \to -\infty} f(x)$ = $\frac{\sqrt{16x^2 - x^2}}{6x^3 + x^2}$

$\frac{\sqrt{16x^6}-\sqrt{x^2}}{6x^3 + x^2}$

$\frac{4x^3-\sqrt{x^2}}{6x^3+x^2}$

$\frac{4-\sqrt{x^2}}{6+x^2}$

My request is may I confirm that I have this correct up to this point?

I do know the final answer, I know the suggested process for calculating the answer, but I want to check that I can also get the correct answer this way. Thank you. Hopefully the formulas come out alright, as I am using Laytex for the first time.

 

[fresh_42]

I have corrected the delimiter $ to ## which is what we use for inline LaTeX code here. I have not corrected what looks like a typo in your "relevant equation" line.

I assume we have a function $f(x)=\dfrac{\sqrt{16x^6-x^2}}{6x^3+x^2}$ and you want to know what $\lim_{x \to -\infty}f(x)$ is.

Note that in general $\sqrt{a+b}\neq \sqrt{a}+\sqrt{b}$. Furthermore, $\sqrt{x^6}=|x^3|=|x|^3$ because we only have the positive square root here. $\sqrt{x^6}=+\sqrt{x^6}=|x|^3.$ Since $x^3$ can assume negative numbers, whereas the positive square root can not, we have to use the absolute value function here.

Please confirm that I got the function right or correct it if I guessed it wrong.

 

[WWGD]

Notice you can pull out the $x^2$ out of the root in the numerator , end up with an $x^1$ on top , then you have $6x^3+x^2$ in the denominator.
Can you see which of the numerator or denominator will dominate the expression?

 

[njh]

I have corrected the delimiter $ to ## which is what we use for inline LaTeX code here. I have not corrected what looks like a typo in your "relevant equation" line.

I assume we have a function $f(x)=\dfrac{\sqrt{16x^6-x^2}}{6x^3+x^2}$ and you want to know what $\lim_{x \to -\infty}f(x)$ is.

Note that in general $\sqrt{a+b}\neq \sqrt{a}+\sqrt{b}$. Furthermore, $\sqrt{x^6}=|x^3|=|x|^3$ because we only have the positive square root here. $\sqrt{x^6}=+\sqrt{x^6}=|x|^3.$ Since $x^3$ can assume negative numbers, whereas the positive square root can not, we have to use the absolute value function here.

Please confirm that I got the function right or correct it if I guessed it wrong.

Yes, that is the correct function. Thank you for the notation correction, as well as the reminder on addition and subtraction under square roots, which I see would have led me down the wrong path. I see the subsequent pointer from WWGD, so I'll try and answer that next and see where it takes me.

In response to WWGD, The denominator dominates the equation. The reason is that $\sqrt{x^6}=|x|^3$, so the equation (in terms of what dominates) is $\dfrac{x^3-x^2}{x^3+x^2}$

I take the point from fresh_42 that I can not square root the numerator's terms individually' as $\sqrt{a+b}\neq \sqrt{a}+\sqrt{b}$

I have tried to pull out the x^2 from the numerator, though I get 1 and not x, so I have missed something here $\dfrac{\sqrt{16x^6-x^2}}{x^2}=\sqrt{x^2(16x^4-1)}$ I do not see that helps me and it is not what WWGD implied, so I have missed something. I will go away and think about this.

 

[bamboum]

Function equivalent in terms dominating numerator and denominator is 4x^3/6x^3=4/6

 

[Mark44]

Function equivalent in terms dominating numerator and denominator is 4x^3/6x^3=4/6

No, because as already stated, $\sqrt{16x^6} \ne 4x^3$. For example, if x = -1, the left side simplifies to 4, but the right side simplifies to -4.

 

[njh]

Yes, that is the correct function. Thank you for the notation correction, as well as the reminder on addition and subtraction under square roots, which I see would have led me down the wrong path. I see the subsequent pointer from WWGD, so I'll try and answer that next and see where it takes me.

In response to WWGD, The denominator dominates the equation. The reason is that $\sqrt{x^6}=|x|^3$, so the equation (in terms of what dominates) is $\dfrac{x^3-x^2}{x^3+x^2}$

I take the point from fresh_42 that I can not square root the numerator's terms individually' as $\sqrt{a+b}\neq \sqrt{a}+\sqrt{b}$

I have tried to pull out the x^2 from the numerator, though I get 1 and not x, so I have missed something here $\dfrac{\sqrt{16x^6-x^2}}{x^2}=\sqrt{x^2(16x^4-1)}$ I do not see that helps me and it is not what WWGD implied, so I have missed something. I will go away and think about this.

No, because as already stated, $\sqrt{16x^6} \ne 4x^3$. For example, if x = -1, the left side simplifies to 4, but the right side simplifies to -4.

I think bamboum just meant in terms of whether the numerator or denominator dominates the equation, so they are just saying that the denominator dominates, which as a starting indicator sounds correct. Oh, but I think that I see what you mean. Whether it is + or - is not.

 

[fresh_42]

Yes, that is the correct function. Thank you for the notation correction, as well as the reminder on addition and subtraction under square roots, which I see would have led me down the wrong path. I see the subsequent pointer from WWGD, so I'll try and answer that next and see where it takes me.

In response to WWGD, The denominator dominates the equation. The reason is that $\sqrt{x^6}=|x|^3$, so the equation (in terms of what dominates) is $\dfrac{x^3-x^2}{x^3+x^2}$

I take the point from fresh_42 that I can not square root the numerator's terms individually' as $\sqrt{a+b}\neq \sqrt{a}+\sqrt{b}$

I have tried to pull out the x^2 from the numerator, though I get 1 and not x, so I have missed something here $\dfrac{\sqrt{16x^6-x^2}}{x^2}=\sqrt{x^2(16x^4-1)}$ I do not see that helps me and it is not what WWGD implied, so I have missed something. I will go away and think about this.

Proceed step by step. We have $f(x)=\dfrac{\sqrt{16x^6-x^2}}{6x^3+x^2}.$ Don't remove parts of the equation too early. By $\sqrt{a \cdot b}=\sqrt{a}\cdot \sqrt{b}$ we get $f(x)=\dfrac{|x|\sqrt{16x^4-1}}{6x^3+x^2}.$ We can assume that $x<0$ since we are interested in $x\rightarrow -\infty .$ So
\begin{align*}
\lim_{x \to -\infty}f(x)&=\lim_{x \to -\infty}\dfrac{|x|\sqrt{16x^4-1}}{6x^3+x^2}=\lim_{x \to -\infty}\dfrac{\sqrt{16x^4-1}}{-6x^2+|x|}
\end{align*}
Now, @WWGD's hint asks: what happens with decreasing $x,$ i.e. with $x\rightarrow -\infty .$ As you correctly observed, the higher powers of $x$ dominate the lower powers of $x$ which becomes irrelevant by decreasing $x.$ Now we can write
\begin{align*}
\lim_{x \to -\infty}f(x)&=\lim_{x \to -\infty}\dfrac{\sqrt{16x^4-1}}{-6x^2+|x|}=\lim_{x \to -\infty}\dfrac{\sqrt{16x^4}}{-6x^2}=\ldots
\end{align*}
You have to keep the signs in mind at every step. We have $f(x)=\dfrac{+}{-}=-$ for $x<0.$

 

[njh]

Fantastic, so -2/3.

There were a couple of things that I hit a wall on.
The first was watching the negatives at each stage, which with hindsight just requires care.
The second was that I initially tried to get rid of the square root as a way to simplify, which proved wrong.
The one thing that surprised me is the ignoring of |x|. I take your point that the dominating factor is the highest power, but I thought that I still had to get rid of any other value for x, which I now is not necessary. Ignoring something close to infinity is counter intuitive.

Thank you for that additional direction. As I had not understood that last principle, I would not have got any further alone.