This Limit problem seems too simple....

[Wi_N]

Homework Statement

$$\lim_{x\to\infty} (x^3 +x^2 +\frac{x}{2})e^{\frac{1}{x}} - \sqrt{x^6+1}$$

Relevant Equations

$$e^{\frac{1}{x}}=1, x\xrightarrow{}\infty$$

$$\lim_{x\to\infty} (x^3+x^2 +\frac{x}{2})-x^3\sqrt{(1-\frac{1}{x^6})} = \lim_{x\to\infty} x^3+x^2+ \frac{x}{2} -x^3=\lim_{x\to\infty} x^2 + \frac{x}{2} = \infty. $$

is this it?

 

[member 587159]

No. You can't do what you did. You cannot evaluate one part of the limit and let the other one stay. You have to take the limit of everything at the same time (a bit hard to explain why this is now allowed).

 

[fresh_42]

I think you must be more careful: the term $e^{1/x}$ makes $x^3$ a bit larger as well as $x^6+1$ is a bit larger. But they do not increase equally, so will one outnumber the other? I would require a series expansion of both, the exponential function as well as the root.

 

[Wi_N]

I think you must be more careful: the term $e^{1/x}$ makes $x^3$ a bit larger as well as $x^6+1$ is a bit larger. But they do not increase equally, so will one outnumber the other? I would require a series expansion of both, the exponential function as well as the root.

expansion like taylor? L'hospital has no use here?

No. You can't do what you did. You cannot evaluate one part of the limit and let the other one stay. You have to take the limit of everything at the same time (a bit hard to explain why this is now allowed).

but i can divide them? lim (x^3...) - lim sqrt(x^6...)

 

[fresh_42]

Laurent, but yes. Substitute $u=1/x$ and use the normal Taylor. If it is infinite, you could as well use a sequence which is lower and infinite and prove that: an estimation from below.

 

[Wi_N]

Laurent, but yes. Substitute $u=1/x$ and use the normal Taylor. If it is infinite, you could as well use a sequence which is lower and infinite and prove that: an estimation from below.

this is calc 101... how many terms of e^ u taylor should i use?

 

[fresh_42]

Then show that $(x^3+x^2+\dfrac{x}{2})e^{\frac{1}{x}} -\sqrt{x^6+1} > x^2$ for large $x$.

 

[Wi_N]

Then show that $(x^3+x^2+\dfrac{x}{2})e^{\frac{1}{x}} -\sqrt{x^6+1} > x^2$ for large $x$.

why not just multiple e^1/x into the bracket and convert all x->infinity at once?

 

[fresh_42]

Let me put it this way:
\begin{align*}
\lim_{x\to \infty}\left[(x^3+x^2+\dfrac{x}{2})e^{1/x}-\sqrt{x^6+1}\right]&=\lim_{x\to \infty}\left[x^3e^{1/x}-x^3\sqrt{1+\frac{1}{x^6}}\right]\\
= \lim_{x\to \infty} \left[x^3\left(e^{1/x}-\sqrt{1+\dfrac{1}{x^6}}\right)\right]\\
=\lim_{x\to \infty} \left(x^3 \cdot 0 \right)\\
= 0
\end{align*}
This is wrong, but why?
Or is it $\ldots = \lim_{x\to \infty}(x^3\cdot (1-1-\sqrt{\dfrac{1}{x^6}}))= -1$, or is it $\ldots = \lim_{x\to \infty}(x^3\cdot (1-\sqrt{1+\dfrac{1}{x^6}}))= -\infty\,?$ These are wrong, too, but why? And why did you get $x^2\,?$ In fact it is $2x^2$.

 

[Wi_N]

i didn't get x^2...what are you talking about? the answer is infinite.

 

[WWGD]

Homework Statement:: $$\lim_{x\to\infty} (x^3 +x^2 +\frac{x}{2})e^{\frac{1}{x}} - \sqrt{x^6+1}$$
Homework Equations:: $$e^{\frac{1}{x}}=1, x\xrightarrow{}\infty$$

$$\lim_{x\to\infty} (x^3+x^2 +\frac{x}{2})-x^3\sqrt{(1-\frac{1}{x^6})} = \lim_{x\to\infty} x^3+x^2+ \frac{x}{2} -x^3=\lim_{x\to\infty} x^2 + \frac{x}{2} = \infty. $$

is this it?

Maybe you can express the product with $e^{1/x}$ as a ratio , an indeterminate form and apply L'Hopital.

 

[Wi_N]

Maybe you can express the product with $e^{1/x}$ as a ratio , an indeterminate form and apply L'Hopital.

that is what I am thinking...thanks.

 

[fresh_42]

i didn't get x^2...what are you talking about? the answer is infinite.

$\ldots =\lim_{x\to\infty} x^2 + \frac{x}{2} = \infty.$

is this it?

You concluded that the limit goes with $x^2+\ldots$. It does with $2x^2+\ldots$. But anyway, that's not the point. Your proof has the same flaws as what I have written in the previous post. So why is yours correct and my versions are wrong?

To solve this dilemma you could either use series expansions or the majority criterion:
$(x^3+x^2+\dfrac{x}{2})e^{1/x}-\sqrt{x^6+1} > x^3e^{1/x} - \sqrt{2x^6} = x^3(e^{1/x}-\sqrt{2}) > \ldots$
or similar.

 

[Wi_N]

You concluded that the limit goes with $x^2+\ldots$. It does with $2x^2+\ldots$. But anyway, that's no the point. Your proof has the same flaws as what I have written in the previous post. So why is yours correct and my versions are wrong?

To solve this dilemma you could either use series expansions or the majority criterion:
$(x^3+x^2+\dfrac{x}{2})e^{1/x}-\sqrt{x^6+1} > x^3e^{1/x} - \sqrt{2x^6} = x^3(e^{1/x}-\sqrt{2}) > \ldots$
or similar.

thanks i think this is a bit advanced for calc 101. ill try hospital instead. what do you think about that?

 

[fresh_42]

thanks i think this is a bit advanced for calc 101. ill try hospital instead. what do you think about that?

Could work. I haven't checked, since L'Hôpital didn't look very nice to me. I think finding a function which is less than the given one at each point and still diverges seems to be easier.

 

[archaic]

This problem can be resolved with the fact that the limit of a polynomial $P(x)$ as $x\to\infty$ is equal to the limit of its term of highest degree and then using l'Hôpital's rule, and I think this is what is looked for in a calculus 1 course.
$$
L=\lim_{x\to\infty} (x^3 +x^2 +\frac{x}{2})e^{\frac{1}{x}} - \sqrt{x^6+1}
=\lim_{x\to\infty}(x^3+x^2+\frac{x}{2})\lim_{x\to\infty}e^\frac{1}{x}-\sqrt{\lim_{x\to\infty}x^6+1}
$$You can replace $\lim_{x\to\infty}P(x)$ by $\lim_{x\to\infty}cx^n$ where $cx^n$ is the term of highest degree.
$$\begin{align*}
L&=\lim_{x\to\infty}(x^3+x^2+\frac{x}{2})\lim_{x\to\infty}e^\frac{1}{x}-\sqrt{\lim_{x\to\infty}x^6+1}\\
&=\lim_{x\to\infty}x^3\lim_{x\to\infty}e^\frac{1}{x}-\sqrt{\lim_{x\to\infty}x^6}\\
&=\lim_{x\to\infty}x^3e^\frac{1}{x}-\lim_{x\to\infty}|x^3|\\
&=\lim_{x\to\infty}x^3e^\frac{1}{x}-x^3\text{ since }x>0\\
&=\lim_{x\to\infty}x^3(e^\frac{1}{x}-1)\sim\infty\cdot0\\
\end{align*}$$

 

[StoneTemplePython]

Then show that $(x^3+x^2+\dfrac{x}{2})e^{\frac{1}{x}} -\sqrt{x^6+1} > x^2$ for large $x$.

further on this point, for any $x \geq 0$
$\sqrt{x^6+1}= \sqrt{(x^3)^2+1^2}\leq \sqrt{(x^3)^2}+\sqrt{1^2} = x^3 + 1$
this is triangle inequality... negating then gives the 'desired direction'

learning to apply triangle inequality seems far more important than blasting this with l'hopital, though I suppose I don't know what is actually emphasized in calc 101

 

[WWGD]

further on this point, for any $x \geq 0$
$\sqrt{x^6+1}= \sqrt{(x^3)^2+1^2}\leq \sqrt{(x^3)^2}+\sqrt{1^2} = x^3 + 1$
this is triangle inequality... negating then gives the 'desired direction'

learning to apply triangle inequality seems far more important than blasting this with l'hopital, though I suppose I don't know what is actually emphasized in calc 101

If the limit goes to $\infty$ that 1 won't matter a bit, triangle inequality or not. So knowing where the limit is headed does matter in the end.

 

[fresh_42]

If the limit goes to $\infty$ that 1 won't matter a bit, triangle inequality or not.

No, not at all. Estimations from above and from below are standard techniques in many parts of mathematics and physics. L'Hôpital is something to look up. And what's even more important: it is far easier! Finding the shortest way is also something to practice.

So knowing where the limit is headed does matter in the end.

Yep. WolframAlpha does it in a second, why bother math? Where the limit is heading is important.

And next week we speak about asymptotes which the same recycled limit, because this week only where the limit is heading is important.

 

[WWGD]

No, not at all. Estimations from above and from below are standard techniques in many parts of mathematics and physics. L'Hôpital is something to look up. And what's even more important: it is far easier! Finding the shortest way is also something to practice.

Yep. WolframAlpha does it in a second, why bother math? Where the limit is heading is important.

And next week we speak about asymptotes which the same recycled limit, because this week only where the limit is heading is important.

How will an extra 1 matter if your limit is $\infty$?

 

[fresh_42]

How will an extra 1 matter if your limit is $\infty$?

It is only a method to get rid of the disturbing $+1$ in the radical. But the main reason I prefer the approximation over L'Hôpital (one more and I program a key with that name ) is, that students tend to choose the narrowest approximations possible whence any rough calculation will do. You can see this in continuity and limit calculations when the $N(\varepsilon)$ becomes a monster expression rather than twice of it.

 

[fresh_42]

And $\min(x_i) \leq \overline{x}_{harm} \leq \overline{x}_{geom}\leq \overline{x}_{arithm}\leq \max(x_i)$ as well as $a^2+b^2\leq a^2+2ab+b^2\; , \;a,b>0$ cannot be stressed enough.

 

[WWGD]

And $\min(x_i) \leq \overline{x}_{harm} \leq \overline{x}_{geom}\leq \overline{x}_{arithm}\leq \max(x_i)$ as well as $a^2+b^2\leq a^2+2ab+b^2\; , \;a,b>0$ cannot be stressed enough.

But this is Calc 101. You're going overboard.

 

[Wi_N]

This problem can be resolved with the fact that the limit of a polynomial $P(x)$ as $x\to\infty$ is equal to the limit of its term of highest degree and then using l'Hôpital's rule, and I think this is what is looked for in a calculus 1 course.
$$
L=\lim_{x\to\infty} (x^3 +x^2 +\frac{x}{2})e^{\frac{1}{x}} - \sqrt{x^6+1}
=\lim_{x\to\infty}(x^3+x^2+\frac{x}{2})\lim_{x\to\infty}e^\frac{1}{x}-\sqrt{\lim_{x\to\infty}x^6+1}
$$You can replace $\lim_{x\to\infty}P(x)$ by $\lim_{x\to\infty}cx^n$ where $cx^n$ is the term of highest degree.
$$\begin{align*}
L&=\lim_{x\to\infty}(x^3+x^2+\frac{x}{2})\lim_{x\to\infty}e^\frac{1}{x}-\sqrt{\lim_{x\to\infty}x^6+1}\\
&=\lim_{x\to\infty}x^3\lim_{x\to\infty}e^\frac{1}{x}-\sqrt{\lim_{x\to\infty}x^6}\\
&=\lim_{x\to\infty}x^3e^\frac{1}{x}-\lim_{x\to\infty}|x^3|\\
&=\lim_{x\to\infty}x^3e^\frac{1}{x}-x^3\text{ since }x>0\\
&=\lim_{x\to\infty}x^3(e^\frac{1}{x}-1)\sim\infty\cdot0\\
\end{align*}$$

the answer is not 0 it goes to infinity.

 

[archaic]

the answer is not 0 it goes to infinity.

$\infty\cdot0$ is an indeterminate form, not $0$! But with that simplified expression you can use l'Hôpital's rule. Try to transform the $\infty\cdot0$ into either $\frac{\infty}{\infty}$ or $\frac{0}{0}$ form (chose the transformation that will let you differentiate easily of course).
Hint: $a=\frac{1}{\frac{1}{a}}$.

 

[WWGD]

6n

further on this point, for any $x \geq 0$
$\sqrt{x^6+1}= \sqrt{(x^3)^2+1^2}\leq \sqrt{(x^3)^2}+\sqrt{1^2} = x^3 + 1$
this is triangle inequality... negating then gives the 'desired direction'

learning to apply triangle inequality seems far more important than blasting this with l'hopital, though I suppose I don't know what is actually emphasized in calc 101

This is Calc 101. I doubt they study triangle inequality.

 

[Wi_N]

$\infty\cdot0$ is an indeterminate form, not $0$! But with that simplified expression you can use l'Hôpital's rule. Try to transform the $\infty\cdot0$ into either $\frac{\infty}{\infty}$ or $\frac{0}{0}$ form (chose the transformation that will let you differentiate easily of course).
Hint: $a=\frac{1}{\frac{1}{a}}$.

using Lhospital now yields no results.

 

[archaic]

using Lhospital now yields no results.

$$\lim_{x\to\infty}x^3(e^\frac{1}{x}-1)
=\lim_{x\to\infty}\frac{e^\frac{1}{x}-1}{\frac{1}{x^3}}
=\lim_{x\to\infty}\frac{-\frac{1}{x^2}e^\frac{1}{x}}{-\frac{3}{x^4}}
=\lim_{x\to\infty}\frac{-x^4}{-3x^2}e^\frac{1}{x}
$$

 

[Wi_N]

$$\lim_{x\to\infty}x^3(e^\frac{1}{x}-1)
=\lim_{x\to\infty}\frac{e^\frac{1}{x}-1}{\frac{1}{x^3}}
=\lim_{x\to\infty}\frac{-\frac{1}{x^2}e^\frac{1}{x}}{-\frac{3}{x^4}}
=\lim_{x\to\infty}\frac{-x^4}{-3x^2}e^\frac{1}{x}
$$

how many times do i have to take the derivative?

 

[archaic]

how many times do i have to take the derivative?

We only need to do it once then simplify the $x$s in this case (remember that $a^na^m=a^{n+m}$ and that $\frac{1}{a^n}=a^{-n}$).
$$\lim_{x\to\infty}\frac{-x^4}{-3x^2}e^\frac{1}{x}=\lim_{x\to\infty}\frac{x^2}{3}e^\frac{1}{x}=\infty$$