Undergrad Finding an infinitesimal limit function

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The discussion focuses on finding an analytical solution for a function as h approaches zero, specifically the limit of the expression involving square roots. The user successfully calculates the limit for the first function, arriving at the result of x divided by the square root of x squared plus y squared. They seek guidance on reducing a second related equation, suggesting that a similar approach involving the difference of square roots could be effective. The technique of multiplying by the conjugate is emphasized as a useful method for evaluating limits. Overall, the conversation centers on limit evaluation techniques for functions involving square roots.
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TL;DR
A function which is the infinitesimal limit of another function
I have the function: ##\sqrt{\left(\frac{x}{h}+1\right)^{2}+\left(\frac{y}{h}\right)^{2}}-\sqrt{\left(\frac{x}{h}\right)^{2}+\left(\frac{y}{h}\right)^{2}}##

I would like to find an analytical solution, the equivalent function, in the limit of h approaching zero.Additional info which might be useful, there's the function: ##\frac{1}{k}\left(\sqrt{\frac{\left(\left(1-k^{2}\right)\cdot y\right)^{2}}{x^{2}+\left(1-k^{2}\right)\cdot y^{2}}+\left(\frac{x}{\sqrt{x^{2}+\left(1-k^{2}\right)\cdot y^{2}}}+k\right)^{2}}-1\right)##

I believe that it also approaches the same solution function in the limit of k approaching zero.
 
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You can calculate the limit by taking out a factor of ##\frac 1 t##, then using the usual technique of multiplying by the conjugate ##\sqrt X + \sqrt Y## etc.
 
$$\sqrt{a}-\sqrt{b} = \frac{a-b}{\sqrt{a}+\sqrt{b}}$$
The right hand side will make it easier to evaluate the limit.
 
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mfb said:
$$\sqrt{a}-\sqrt{b} = \frac{a-b}{\sqrt{a}+\sqrt{b}}$$
The right hand side will make it easier to evaluate the limit.

That's what I meant!
 
awesome, I was able to calculate it, ##\frac{x}{\sqrt{x^{2}+y^{2}}}##
 
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how do you think I can reduce the 2nd equation?
 
The same approach should work there. ##1=\sqrt{1}## and you have a difference of square roots again.
 

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