Help Taking the Limit as K goes to infinity

[YoshiMoshi]

Homework Statement

Evaluate the limit as K goes to infinity of s_1,2 (K)

Homework Equations

The Attempt at a Solution

Apparently my value for plus the square root is incorrect, apparently the correct answer is 1.

Apparently my value for minus the square root is correct, it's negative infinity.

I'm not sure what I'm doing wrong. I see when I take plus the stuff in the square root.

-5/2 - infinity + infinity = - 5/2

Apparently this is wrong.

I used MATLAB and got plus 1.

 

[gneill]

Looks to me that you have a situation where l'Hopital's Rule might come to your rescue. You have a situation where you're taking the difference between two infinities. Try to cast it into the form of a quotient of infinities and apply l'Hopital.

Hint: For the moment ignore the part that is the simple constant -5/2 and examine the rest. Divide numerator and denominator by K to form your f(x) and g(x) for l'Hopital. Yes, derivatives involving square roots are annoying, but what can you do?

 

[YoshiMoshi]

I thought L'Hopital's Rule was for

0/0, or infinity/infinity

by L'Hopital's Rule, I take the derivative of numerator divided by the derivative of the denominator

But I can apply L'Hopital's Rule in this case as well where I don't have the form 0/0, infinity/infinity?

 

[Delta2]

$-\infty+\infty$ does not equal zero, it is undefined. You must treat the expression differently in this case.
Try to write

$$-K+\sqrt{K^2+14K+1}=Kf(K)$$,

that is to factor out K from the whole expression and see what is $f(K)$. It will be $\lim_{K\rightarrow +\infty}f(K)=0$ so again you will have an undefined form of $+\infty\cdot 0$ and you ll have to use L'Hopital rule .

 

[gneill]

I thought L'Hopital's Rule was for

0/0, or infinity/infinity

by L'Hopital's Rule, I take the derivative of numerator divided by the derivative of the denominator

But I can apply L'Hopital's Rule in this case as well where I don't have the form 0/0, infinity/infinity?

Yes, if you cast the 'difficult' part of the equation into the form f(x)/g(x) and both f(x) and g(x) tend to 0 or infinity.

 

[YoshiMoshi]

Thanks for the help. Following the guidance I got to this point

Seems that I would get infinity/infinity when taking the limit because of this term, so I need to apply the L Hospital's rule again?

 

[Delta2]

No , the limit of that term is simply 1. (factor out K from both numerator and denominator).
But two things :

1) You should keep the 5/2, its not part of the L Hopital rule, so you don't take the derivative of it to make it zero...
2) You have at some point $\frac{5}{2}+\frac{-1+\frac{\sqrt{K^2+14K+1}}{K}}{\frac{2}{K}}$. Try to first simplify the term $\frac{\sqrt{K^2+14K+1}}{K}$ and then apply L'Hopital rule.

Also the "-" case is covered , we know that the limit is $-\infty$, we are interested for the "+" case here.

 

[Mark44]

@YoshiMoshi, some of your work in post #1 is just plain wrong -- for instance where you wrote $\frac \infty 2$ and $\sqrt{\infty^2 + 14\infty + 1}$. You cannot use $\infty$ in any arithmetic expression as you did there.
@Delta2, L'Hopital's Rule is not needed here.

In the original problem, one of the two equations is $s_2(k) = -\frac 5 2 + \frac 1 2(-k + \sqrt{k^2 + 14k + 1})$. The other limit is straightforward, so I'm not considering it here.

To make life simpler, work with $\lim_{k \to \infty}(-k + \sqrt{k^2 + 14k + 1})$. You can evaluate this limit by multplying by 1, in the form of $-k - \sqrt{k^2 + 14k + 1}$ over itself. When you do this, it's just a matter of algebraic manipulation.

After you get a limit for this expression, multiiply that result by 1/2 and then add -5/2. I get the same value you got from matlab.

 

[Delta2]

@Mark44 You are right it can be done easier without L'Hopital. I worked it out anyway with one application of L'Hopital and found it equal to $-5/2+14/4=1$.

 

[Buzz Bloom]

Apparently this is wrong.

Hi Yoshi:

I think a simpler approach would be to factor the square root into the following form:
√(k² +14k+1) = k√(1+...).

Hope that helps.

BTW: Sorry - I am now correcting an careless error. The correct result is 1.
You will need to find an approximate square root of (1+...)

Regards,
Buzz

 

[Mark44]

You will need to find an approximate square root of (1+...)

Since we're taking the limit as $k \to \infty$, we don't need to approximate the square root.

 

[Buzz Bloom]

Since we're taking the limit as k→∞k \to \infty, we don't need to approximate the square root.

Hi Mark:

I guess I was not clear about "approximation".

√(k² +14k+1) = k√(1+ 14/k + 1/k²)

= k (1+7/k + (1/2)/k² + O(1/k³)) = k + 7 + 1/2k + O(1/k²)) .​

The 1/2k term can be ignored since it goes to 0 as k goes to infinity.

Regards,
Buzz

 

[Mark44]

Hi Mark:

I guess I was not clear about "approximation".

√(k² +14k+1) = k√(1+ 14/k + 1/k²)

= k (1+7/k + (1/2)/k² + O(1/k³)) = k + 7 + 1/2k + O(1/k²)) .​

The 1/2k term can be ignored since it goes to 0 as k goes to infinity.

My only objection to doing things this way, with a truncated binomial series, is that the OP might not have been exposed to series yet. To my way of thinking, it's much simpler to write $\sqrt{k^2 + 14k + 1}$ as $k\sqrt{1 + \frac {14} k + \frac 1 {k^2}}$.
Evaluating this limit, together with the rest of the original problem, is pretty straightforward.

 

[Buzz Bloom]

the OP might not have been exposed to series yet.

Hi Mark:

You may well be correct about what the OP has previously been exposed to. My thought was that he might have been exposed to the process of calculating an approximate value of of √(1+ε) since he is able to take derivatives. I have no idea how math is taught currently, but some while ago calculus would come after dealing with infinitesimals.

Regards,
Buzz

 

[epenguin]

This is two questions because of the ± .
I think the - case is obvious and not very interesting.

For the + case I think what they are looking for is: express the quadratic under the square root as the sum of a perfect square plus a constant (which may be positive or negative). As K "goes to infinity" or "becomes as large as we please" it will approximate ever closer to the square root of the perfect square - that is to the thing it is the square of. (You can easily convince yourself of this assertion with some numerical examples if you have a calculator or computer that can do square root use the biggest numbers it can handle, e.g. K equals 1 million and 100 million. You need also to justify it more than that). I got the answer 1 very easily by the method outlined.

When I see problems like this I doubt that they come out of the blue. I would bet that if you look back in your textbook you will find worked examples of something similar.