This Limit problem seems too simple....

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In summary: R},\text{ so }|x^3|=x^3\\&=\infty\end{align*}$$In summary, we can solve this problem by using the fact that the limit of a polynomial as x approaches infinity is equal to the limit of its term of highest degree, and applying L'Hôpital's rule. We can then use the fact that, for all real numbers, x^3 is greater than 0, to simplify the equation and ultimately conclude that the limit is equal to infinity.
  • #1
Wi_N
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Homework Statement
$$\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?
 
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  • #2
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).
 
  • #3
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.
 
  • #4
fresh_42 said:
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?

Math_QED said:
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...)
 
  • #5
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.
 
  • #6
fresh_42 said:
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?
 
  • #7
Then show that ##(x^3+x^2+\dfrac{x}{2})e^{\frac{1}{x}} -\sqrt{x^6+1} > x^2## for large ##x##.
 
  • #8
fresh_42 said:
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?
 
  • #9
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##.
 
  • #10
i didn't get x^2...what are you talking about? the answer is infinite.
 
  • #11
Wi_N said:
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.
 
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  • #12
WWGD said:
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.
 
  • #13
Wi_N said:
i didn't get x^2...what are you talking about? the answer is infinite.
Wi_N said:
##\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.
 
  • #14
fresh_42 said:
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?
 
  • #15
Wi_N said:
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.
 
  • #16
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.
$$\begin{align*}
L&=\lim_{x\to\infty}(x^3+x^2+\frac{x}{2})\lim_{x\to\infty}e^\frac{1}{x}-\sqrt{\lim_{x\to\infty}x^6+1}\\
&=\lim_{x\to\infty}x^3\lim_{x\to\infty}e^\frac{1}{x}-\sqrt{\lim_{x\to\infty}x^6}\\
&=\lim_{x\to\infty}x^3e^\frac{1}{x}-\lim_{x\to\infty}|x^3|\\
&=\lim_{x\to\infty}x^3e^\frac{1}{x}-x^3\text{ since }x>0\\
&=\lim_{x\to\infty}x^3(e^\frac{1}{x}-1)\sim\infty\cdot0\\
\end{align*}$$
 
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  • #17
fresh_42 said:
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
 
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  • #18
StoneTemplePython said:
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.
 
  • #19
WWGD said:
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.
 
  • #20
fresh_42 said:
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##?
 
  • #21
WWGD said:
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 :mad:) 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.
 
  • #22
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.
 
  • #23
fresh_42 said:
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.
 
  • #24
archaic said:
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.
 
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  • #25
Wi_N said:
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}}##.
 
  • #26
6n
StoneTemplePython said:
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.
 
  • #27
archaic said:
##\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.
 
  • #28
Wi_N said:
using Lhospital now yields no results.
$$\lim_{x\to\infty}x^3(e^\frac{1}{x}-1)
=\lim_{x\to\infty}\frac{e^\frac{1}{x}-1}{\frac{1}{x^3}}
=\lim_{x\to\infty}\frac{-\frac{1}{x^2}e^\frac{1}{x}}{-\frac{3}{x^4}}
=\lim_{x\to\infty}\frac{-x^4}{-3x^2}e^\frac{1}{x}
$$
 
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  • #29
archaic said:
$$\lim_{x\to\infty}x^3(e^\frac{1}{x}-1)
=\lim_{x\to\infty}\frac{e^\frac{1}{x}-1}{\frac{1}{x^3}}
=\lim_{x\to\infty}\frac{-\frac{1}{x^2}e^\frac{1}{x}}{-\frac{3}{x^4}}
=\lim_{x\to\infty}\frac{-x^4}{-3x^2}e^\frac{1}{x}
$$
how many times do i have to take the derivative?
 
  • #30
Wi_N said:
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}##).
$$\lim_{x\to\infty}\frac{-x^4}{-3x^2}e^\frac{1}{x}=\lim_{x\to\infty}\frac{x^2}{3}e^\frac{1}{x}=\infty$$
 
Last edited:
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  • #31
StoneTemplePython said:
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

You don't even need the triangle inequality. Simply:

##(x^3 + 1)^2 = x^6 + 2x^3 + 1 > x^6 + 1 \ \ (x > 0)##

##x^3 + 1 > \sqrt{x^6+1}##
 
  • #32
archaic said:
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}##).
$$\lim_{x\to\infty}\frac{-x^4}{-3x^2}e^\frac{1}{x}=\lim_{x\to\infty}\frac{x^2}{3}e^\frac{1}{x}=\infty$$

is there a specific reason u kept the higher terms and removed the lower terms? is there a reasoning?
 
  • #33
Wi_N said:
is there a specific reason u kept the higher terms and removed the lower terms? is there a reasoning?

Let ##f(x) = g(x) - h(x)##

Suppose we can show that ##g(x) - h(x)## diverges to ##+\infty##, then we know that ##f(x)## diverges.

Now, suppose we find ##g_2(x) < g(x)##, then:

##f(x) > g_2(x) - h(x)##

And, if we can see that ##g_2(x) - h(x)## diverges to ##+\infty##, then we know that ##f(x)## diverges.

Finally, if we can find ##h_2(x) > h(x)##, then:

##f(x) > g_2(x) - h_2(x)##

And, if we can see that ##g_2(x) - h_2(x)## diverges to ##+\infty##, then we know that ##f(x)## diverges.

That's the idea.
 
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  • #34
Wi_N said:
is there a specific reason u kept the higher terms and removed the lower terms? is there a reasoning?
I do not think that this is rigorous but it is intuitive.
$$
\lim_{x\to\infty}\frac{2x^3+x+6}{x^3}
=\lim_{x\to\infty}\frac{x^3(2+\underbrace{\frac{x}{x^3}}_\text{goes to 0}+\underbrace{\frac{6}{x^3}}_\text{goes to 0})}{x^3}=2\\
\Leftrightarrow \lim_{x\to\infty}\frac{2x^3+x+6}{x^3}=2\\
\Leftrightarrow \frac{\lim_{x\to\infty}2x^3+x+6}{\lim_{x\to\infty}x^3}=2\\
\Leftrightarrow \lim_{x\to\infty}2x^3+x+6=2\lim_{x\to\infty}x^3\\
\Leftrightarrow \lim_{x\to\infty}2x^3+x+6=\lim_{x\to\infty}2x^3
$$
 
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  • #35
archaic said:
I do not think that this is rigorous but it is intuitive.
$$
\lim_{x\to\infty}\frac{2x^3+x+6}{x^3}
=\lim_{x\to\infty}\frac{x^3(2+\underbrace{\frac{x}{x^3}}_\text{goes to 0}+\underbrace{\frac{6}{x^3}}_\text{goes to 0})}{x^3}=2\\
\Leftrightarrow \lim_{x\to\infty}\frac{2x^3+x+6}{x^3}=2\\
\Leftrightarrow \frac{\lim_{x\to\infty}2x^3+x+6}{\lim_{x\to\infty}x^3}=2\\
\Leftrightarrow \lim_{x\to\infty}2x^3+x+6=2\lim_{x\to\infty}x^3\\
\Leftrightarrow \lim_{x\to\infty}2x^3+x+6=\lim_{x\to\infty}2x^3
$$

In general, this doesn't hold. Especially:
$$
\lim_{x\to\infty}\frac{2x^3+x+6}{x^3}=2\\
\Rightarrow \frac{\lim_{x\to\infty}2x^3+x+6}{\lim_{x\to\infty}x^3}=2\\
$$
is false.
 
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<h2>What is a limit problem?</h2><p>A limit problem is a mathematical concept that involves finding the value that a function approaches as the input approaches a certain value. It is used to describe the behavior of a function near a particular point.</p><h2>Why does this limit problem seem too simple?</h2><p>This limit problem may seem too simple because it involves a basic function and a straightforward input value. However, even simple limit problems can have complex solutions and can be used to understand more complicated functions.</p><h2>What is the importance of solving limit problems?</h2><p>Solving limit problems is important because it helps us understand the behavior of functions and their values at specific points. This concept is essential in many areas of mathematics, physics, and engineering.</p><h2>What are some common strategies for solving limit problems?</h2><p>Some common strategies for solving limit problems include using algebraic manipulation, substitution, and L'Hopital's rule. It is also helpful to understand the properties of limits and the different types of indeterminate forms.</p><h2>How can limit problems be applied in real-world situations?</h2><p>Limit problems can be applied in real-world situations to model and predict the behavior of various systems. For example, they can be used to analyze the growth of populations, the speed of moving objects, and the efficiency of chemical reactions.</p>

What is a limit problem?

A limit problem is a mathematical concept that involves finding the value that a function approaches as the input approaches a certain value. It is used to describe the behavior of a function near a particular point.

Why does this limit problem seem too simple?

This limit problem may seem too simple because it involves a basic function and a straightforward input value. However, even simple limit problems can have complex solutions and can be used to understand more complicated functions.

What is the importance of solving limit problems?

Solving limit problems is important because it helps us understand the behavior of functions and their values at specific points. This concept is essential in many areas of mathematics, physics, and engineering.

What are some common strategies for solving limit problems?

Some common strategies for solving limit problems include using algebraic manipulation, substitution, and L'Hopital's rule. It is also helpful to understand the properties of limits and the different types of indeterminate forms.

How can limit problems be applied in real-world situations?

Limit problems can be applied in real-world situations to model and predict the behavior of various systems. For example, they can be used to analyze the growth of populations, the speed of moving objects, and the efficiency of chemical reactions.

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