Limit problem simplify the root

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The discussion revolves around simplifying a mathematical problem where the solution is known to be g/8. Participants express frustration over not being able to manipulate the equation effectively, particularly regarding the square root and its implications. They suggest using the conjugate to aid in simplification and explore factoring out common elements under the square root. The conversation also touches on applying limits and dividing by the highest power of c to simplify the expression further. Ultimately, the contributors help each other reach a clearer understanding of how to approach the problem.
StrSpeed
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Homework Statement


[/B]
I feel like I'm missing some theorem which is preventing me from finalizing this problem! It's been driving me nuts I feel like I'm missing something super basic!

Ultimately they've given the solution, g/8, so I know this is how I should try to get the equation to look algebraically. But, no matter how I manipulate it I can't get it to reduce.

I feel like this has to do with √x2 = |x| Which then depending on your value of x will give x, or -x. However, I can't simplify the root into a way which will let me make this jump.
upload_2015-6-5_12-25-51.png


Homework Equations


https://www.desmos.com/calculator/hf8poewlvb

The Attempt at a Solution


https://www.desmos.com/calculator/hf8poewlvb
(link to all my reductions)
 
Last edited:
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Try multiplying numerator and denominator by$$
c\sqrt{\left(\frac{c^2}{g^2}+\frac 1 4\right)}+\frac{c^2}{g}$$
 
The conjugate! How could I forget that.. Thank you! Now I have:

PS. g is not squared I wrote it down wrong.
VhM1czk.jpg


Let me see what more I can do.
 
Alternatively: You can take a common factor out of the expression under the square root to obtain <br /> h(c) = A\sqrt{1 + x} - \frac{c^2}{g} where x &lt; 1 for sufficiently large c. Hence you may expand the root as a binomial series, <br /> (1 + x)^{\alpha} = 1 + \alpha x + \frac{\alpha(\alpha - 1)}{2!}x^2 + \dots
 
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Thanks for both of your help! Still working on it, sadly.. I pulled out a C so I'm left with:

https://www.physicsforums.com/file:///C:/Users/Steven/Downloads/CodeCogsEqn.gif
7ScJ3py.gif


Im not entirely sure what I could pull out of the root.
 
Last edited by a moderator:
StrSpeed said:
Thanks for both of your help! Still working on it, sadly.. I pulled out a C so I'm left with:

https://www.physicsforums.com/file:///C:/Users/Steven/Downloads/CodeCogsEqn.gif
7ScJ3py.gif


Im not entirely sure what I could pull out of the root.
Factor out the c you'll have 1/4 * (lim c ->0 1/(sqrt(1/g^2 + 1/4c) + 1/g) and don't forget that 1/4c goes to 0 whenever c -> 0, good luck
 
Last edited by a moderator:
StrSpeed said:
The conjugate! How could I forget that.. Thank you! Now I have:

PS. g is not squared I wrote it down wrong.
VhM1czk.jpg


Let me see what more I can do.
That's good.

##\displaystyle \ \frac14\lim_{c\to\infty}\left(\frac{c^2}{\displaystyle c\sqrt{\left(\frac{c^2}{g^2}+\frac 1 4\right)}+\frac{c^2}{g}}\right)\ ##

One way to deal with rational expressions where some factor →∞ : divide the numerator and denominator by the highest power of that factor.

Divide by c2 in the numerator & denominator.
 
I got it! You guys rock thank you so much! Once I got down to that last x Term I realized that a constant/infinity = 0!

Here is my work through!
od4ZL7X.jpg
 
You should drop the limit sign from the second last row.
 
  • #10
StrSpeed said:
I got it! You guys rock thank you so much! Once I got down to that last x Term I realized that a constant/infinity = 0!

Here is my work through!
upload_2015-6-5_22-57-10.png
At the point highlighted above, simply take the limit and simplify.
 

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