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Limit problem

  • Thread starter ciubba
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  • #1
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Homework Statement


Find [tex]lim_{x->- \infty} \; \frac{(x^6+8)^{1/3}}{4x^2+(3x^4+1)^{1/2}}[/tex]

Homework Equations


N/A

The Attempt at a Solution


Factoring out [tex]\frac {(-x^6)^{1/3}}{-x^2}[/tex] leaves me with [tex]\frac{(-1-8x^{-6})}{-4+(3+x^{-4})^{1/2
}}[/tex] Taking the limit at infinity gives me [tex]\frac{-1}{-4+3^{1/2}}[/tex], which is wrong.
 

Answers and Replies

  • #2
Quantum Defect
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Homework Statement


Find [tex]lim_{x->- \infty} \; \frac{(x^6+8)^{1/3}}{4x^2+(3x^4+1)^{1/2}}[/tex]

Homework Equations


N/A

The Attempt at a Solution


Factoring out [tex]\frac {(-x^6)^{1/3}}{-x^2}[/tex] leaves me with [tex]\frac{(-1-8x^{-6})}{-4+(3+x^{-4})^{1/2
}}[/tex] Taking the limit at infinity gives me [tex]\frac{-1}{-4+3^{1/2}}[/tex], which is wrong.
Why not factor out (x^2/x^2) on the top?... i.e. lim x-> -inf [1+8/x^6]^(1/3) / [4 + SQRT(3+ 1/x^4)]
 
  • #3
Svein
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Your arithmetic is wrong (in the denominator).
 
  • #4
Ray Vickson
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Homework Statement


Find [tex]lim_{x->- \infty} \; \frac{(x^6+8)^{1/3}}{4x^2+(3x^4+1)^{1/2}}[/tex]

Homework Equations


N/A

The Attempt at a Solution


Factoring out [tex]\frac {(-x^6)^{1/3}}{-x^2}[/tex] leaves me with [tex]\frac{(-1-8x^{-6})}{-4+(3+x^{-4})^{1/2
}}[/tex] Taking the limit at infinity gives me [tex]\frac{-1}{-4+3^{1/2}}[/tex], which is wrong.
Why not use ##(x^6+8)^{1/3} = (x^6)^{1/3} (1 + 8 x^{-6})^{1/3} \doteq x^2 [1 + (8/3)x^{-6} + O(x^{-12}) ]##, and a similar result for ##(3 x^4 + 1)^{1/2}##?
 
  • #5
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Your arithmetic is wrong (in the denominator).
Ah, you're right. it should be MINUS root 3 because [tex]\sqrt{x^4}=-x^2[/tex]

\doteq x^2 [1 + (8/3)x^{-6} + O(x^{-12}) ]
I don't understand how you arrived at this part.

Why not factor out (x^2/x^2) on the top?... i.e. lim x-> -inf [1+8/x^6]^(1/3) / [4 + SQRT(3+ 1/x^4)]
If I factored out positive x^2, how would I evaluate x at - infinity? Wouldn't I just evaluate at positive infinity, which happens to produce the same result in this case?
 
  • #6
Quantum Defect
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[snip]

If I factored out positive x^2, how would I evaluate x at - infinity? Wouldn't I just evaluate at positive infinity, which happens to produce the same result in this case?
Factoring out +x^2 says nothing about what the sign of x is, does it?

The function whose limit is being evaluated is an even function of x, no? f(-x) = f(x) Wouldn't they be expected to have the same limit for very large positive and very large negative values of x?
 
  • #7
Ray Vickson
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Ah, you're right. it should be MINUS root 3 because [tex]\sqrt{x^4}=-x^2[/tex]


I don't understand how you arrived at this part.

******************************************
Do you agree that ##(x^6+8)^{1/3} = (x^6)^{1/3} (1 + 8 x^{-6})^{1/3} = x^2 (1 + 8 x^{-6})^{1/3}##? If so, then do you agree that we can apply the general binomial expansion? This would be ##(1+v)^r = 1 + rv + O(v^2)##, which holds for ##|v| < 1## and any real ##r## (positive, negative, rational, irrational---anything). If so, just let ##v = 8/x^6## and ##r = 1/3##.
*******************************************

If I factored out positive x^2, how would I evaluate x at - infinity? Wouldn't I just evaluate at positive infinity, which happens to produce the same result in this case?
Ah, you're right. it should be MINUS root 3 because [tex]\sqrt{x^4}=-x^2[/tex]




If I factored out positive x^2, how would I evaluate x at - infinity? Wouldn't I just evaluate at positive infinity, which happens to produce the same result in this case?
 
  • #8
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Can you link me the formula for that expansion? I've never seen it before.
 
  • #10
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Where did the O(v^2) come from? The formula seems to stop at 1+rv.
 
  • #11
Ray Vickson
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Can you link me the formula for that expansion? I've never seen it before.
Google "binomial expansion". Alternatively, just apply the Maclauren (Taylor) series around ##v=0## to the function ##f(v) = (1+v)^r##.
 
  • #12
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2
I won't be learning about power series until the end of next semester, so it's likely that your explanation is a bit above my level for the time being.

I'm still a bit confused by why it's - root 3.

The only example problem in my book defines (x^6)^1/2= - x^3, but when they factor [tex]\sqrt{25x^6}[/tex] it comes out to +5 rather than -5.
 
  • #13
HallsofIvy
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Homework Statement


Find [tex]lim_{x->- \infty} \; \frac{(x^6+8)^{1/3}}{4x^2+(3x^4+1)^{1/2}}[/tex]

Homework Equations


N/A

The Attempt at a Solution


Factoring out [tex]\frac {(-x^6)^{1/3}}{-x^2}[/tex] leaves me with [tex]\frac{(-1-8x^{-6})}{-4+(3+x^{-4})^{1/2
}}[/tex]
No, it doesn't. It leaves you with [tex]\frac{(-1-8x^{-6})}{-4- (3+x^{-4})^{1/2
}}[/tex]

Taking the limit at infinity gives me [tex]\frac{-1}{-4+3^{1/2}}[/tex], which is wrong.
 
  • #14
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  • #15
Ray Vickson
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Why doesn't the root 25 change sign in the following picture: http://s10.postimg.org/qb9mxcsmf/1243124.png

It's the same idea except that instead of [tex]
\sqrt{x^4}=-x^2[/tex] we have [tex]\sqrt{x^6}=-x^3[/tex]
We should NOT have
[tex] \sqrt{x^4} = -x^2 \:\longleftarrow\;\text{wrong!}[/tex]
For any real ##x## we have ##\sqrt{x^4} = x^2## (because ##\sqrt{x^4} > 0 ## and ## x^2 > 0## if ##x \neq 0##; this is the same, whether ##x## is positive or negative. However, ##\sqrt{x^6}## is different. Again, we always have ##\sqrt{x^6} > 0##, but now ##x^3## is >0 if ##x > 0## and is < 0 if ##x < 0##. In fact, if ##x < 0## then ##x^3 < 0##, so we need ##\sqrt{x^6} = -x^3## in that case (where ##x < 0##). On the other hand, if ##x > 0## then we have ##\sqrt{x^6} = x^3##. To summarize:
[tex] \sqrt{x^4} = x^2 \;\; \text{always}\\
\sqrt{x^6} = \begin{cases} x^3 & \text{if} \;\; x > 0 \\
-x^3 & \text{if} \;\; x < 0
\end{cases}
[/tex]
 
Last edited:

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