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Limits and conjugates problem

  • Thread starter ctran
  • Start date
2
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I cannot seem to compute the limit,
lim (sqrt(16x^(2)+15)-4x)/(4x-1000)
x->-inf


Ive tried using L'Hopital's Rule but it made it more confusing and tried using conjugates but that didnt really work either.
1. Homework Statement



2. Homework Equations



3. The Attempt at a Solution
 
251
0
Re: Limits

it goes to 0, after multiplying by the conjugate, you can take the limit of the big expression as x goes to infinity and so the whole big thing in the bottom goes to infinity so you get 0
 
2
0
Re: Limits

Unfortunately this limit does not tend to zero it goes to -2.
 

lanedance

Homework Helper
3,304
2
Re: Limits

[tex] \lim_{x\to-\infty} \frac{\sqrt{16x^2-15}-4x}{4x-1000} [/tex]

how do you get -2?
 
251
0
Re: Limits

it's zero if you aren't sure try graphing it
 
251
0
Re: Limits

a fixed number in the numerator, with the denominator going to infinity.. of course it is zero
 

lanedance

Homework Helper
3,304
2
Re: Limits

actually the negative infinity makes things interesting, try a variable change u = -x, and then multply through by (1/u)/(1/u) and see what happens
 

lanedance

Homework Helper
3,304
2
Re: Limits

i get 8/(-4)? hopefully i didn't miss anything...
 
251
0
Re: Limits

huh, how did you get that

I even graphed it and it looks like the limit is 0, by the way, I never saw the -inf but it didn't matter
 

lanedance

Homework Helper
3,304
2
Re: Limits

i checked on a graph & get -2 as described above... with some real abuse of notation (to avoid doing the whole solution), i think its because:

[tex] \lim_{x\to-\infty} \frac{\sqrt{(16x^2-15)}-4x}{4x-1000} [/tex]

with the variable change u = -x
[tex] \lim_{u\to\infty} \frac{\sqrt{(16u^2-15)}+4u}{-4u-1000} \approx \frac{4(\infty)+ 4u(\infty) }{-4(\infty)}[/tex]

though i agree with you on the positive infinite limit going to 0
 
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251
0
Re: Limits

ah, I see I checked on a graph for positive infinity and assumed that -inf was the same thanks
 
32,773
4,479
Re: Limits

This is much simpler than it would seem from the replies in this thread. All you need to do is to factor 4x from the numerator and denominator. No conjugates, no L'Hopital's rule.
[tex]\frac{\sqrt{16x^2 - 15} - 4x}{4x - 1000}~=~\frac{4x(\sqrt{1 - 15/(16x^2)} - 1)}{4x(1 - 1000/(4x))}~=~\frac{\sqrt{1 - 15/(16x^2)} - 1}{1 - 1000/(4x)}[/tex]

As x approaches negative infinity, the numerator approaches 0 and the denominator approaches 1, making the limit 0.
 

lanedance

Homework Helper
3,304
2
Re: Limits

This is much simpler than it would seem from the replies in this thread. All you need to do is to factor 4x from the numerator and denominator. No conjugates, no L'Hopital's rule.
[tex]\frac{\sqrt{16x^2 - 15} - 4x}{4x - 1000}~=~\frac{4x(\sqrt{1 - 15/(16x^2)} - 1)}{4x(1 - 1000/(4x))}~=~\frac{\sqrt{1 - 15/(16x^2)} - 1}{1 - 1000/(4x)}[/tex]

As x approaches negative infinity, the numerator approaches 0 and the denominator approaches 1, making the limit 0.
i might be missing something, but i'm not convinced about the part factroing 4x out of the square root, i think we might not be preserving the negative and should go something like

[tex] \lim_{x\to-\infty} \frac{\sqrt{(16x^2-15)}-4x}{4x-1000} [/tex]

[tex] \lim_{x\to-\infty} (\frac{1/x}{1/x}) \frac{\sqrt{(16x^2-15)}-4x}{4x-1000} [/tex]

[tex] \lim_{x\to-\infty} \frac{\frac{-1}{\sqrt{x^2}}\sqrt{(16x^2-15)}-4}{4-1000/x} [/tex]

[tex] \lim_{x\to-\infty} \frac{-\sqrt{(16-15/x^2)}-4}{4-1000/x} [/tex]


this is the way i went with the variable change [itex] x = -u, x^2=u^2[/itex]
[tex] \lim_{x\to-\infty} \frac{\sqrt{(16x^2-15)}-4x}{4x-1000} [/tex]

[tex]= \lim_{u\to\infty} \frac{\sqrt{(16u^2-15)}+4u}{-4u-1000} [/tex]

[tex]= \lim_{u\to\infty} (\frac{1/u}{1/u}) \frac{\sqrt{(16u^2-15)}+4u}{-4u-1000} [/tex]

[tex]= \lim_{u\to\infty} \frac{\sqrt{(16-15/u^2)}+4}{-4-1000/u}= \frac{\sqrt{(16)}+4}{-4} = \frac{8}{-4} = -2[/tex]
 
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867
0
Re: Limits

i might be missing something, but i'm not convinced about the part factroing 4x out of the square root, i think we might not be preserving the negative and should gos something like

[tex] \lim_{x\to-\infty} \frac{\sqrt{(16x^2-15)}-4x}{4x-1000} [/tex]

[tex] \lim_{x\to-\infty} (\frac{1/x}{1/x}) \frac{\sqrt{(16x^2-15)}-4x}{4x-1000} [/tex]

[tex] \lim_{x\to-\infty} \frac{\frac{-1}{\sqrt{x^2}}\sqrt{(16x^2-15)}-4}{4-1000/x} [/tex]
Where did the negative sign in -1/√x2 come from in the last line above?

[Edit] I see that you left it out later in your work after the substitution. I agree that the negative isn't preserved when you take x out of the square root and the substitution you use gives the correct limit of -2 (Wolframalpha agrees with you, and it takes care of the negative in an interesting way). Reminds me of finding the derivative of arcsecant...
 
Last edited:

lanedance

Homework Helper
3,304
2
Re: Limits

Where did the negative sign in -1/√x2 come from in the last line above?
from the ether?

but no, being a positive person.... i added it in to preserve the negativity... not 100% sure its legal/rigourous, which is why i prefer the variable change u = -x

So as the limit is heading towards a negative number, when you take that inside the square root, you lose the negativity (ie, 1/x heads to zero from the negative side), so to highlight the fact, i decided to substitute
[tex] \lim_{x\to-\infty} {\frac{1}{x} = \lim_{x\to-\infty} {\frac{-1}{\sqrt{x^2}} [/tex]

though as said the 2nd way of the approaching the problem seems more palatable
 
32,773
4,479
Re: Limits

lanedance, that's a good point about negative numbers that I overlooked. My error was in replacing x^2 inside the radical by x outside it. The actual identity is [itex]\sqrt{x^2}=|x|[/itex].

My revised work follows, and takes into account that the original problem had 16x^2 + 15 under the radical, not 16x^2 + 15 that appeared later.
[tex]\frac{\sqrt{16x^2 + 15} - 4x}{4x - 1000}~=~\frac{4|x|(\sqrt{1 + 15/(16x^2)} + 1)}{-4|x|(1 - 1000/(4x))}~=~\frac{-(\sqrt{1 + 15/(16x^2)} + 1)}{1 - 1000/(4x)}[/tex]

As x approaches negative infinity, the numerator approaches -2 and the denominator approaches 1, so the limit is -2.

A couple of steps above are not obvious, so here's the explanation for them. Since we are taking the limit as x --> -infinity, it's reasonable to assume that x < 0. In that case, x = -|x|, so I can replace -4x by +4|x| in the numerator, and can replace 4x by -4|x| in the denominator. This is why the signs changed in going from the first expression above to the the second.
 

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