How Does the Limit Process Simplify the Expression x^2-36?

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Homework Statement


\lim_{x\to 6}\frac{\sqrt{x+3}-3}{x-6}


Homework Equations



Answer is 1/6.

The Attempt at a Solution


\lim_{x\to 6}\frac{\sqrt{x+3}-3}{x-6}=\lim_{x\to 6}\frac{(x+3-9)(x+6)}{(x^2-36)(\sqrt{x+3}+3)}
 
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fermio said:

Homework Statement


\lim_{x\to 6}\frac{\sqrt{x+3}-3}{x-6}


Homework Equations



Answer is 1/6.

The Attempt at a Solution


\lim_{x\to 6}\frac{\sqrt{x+3}-3}{x-6}=\lim_{x\to 6}\frac{(x+3-9)(x+6)}{(x^2-36)(\sqrt{x+3}+3)}

Look more carefully at your attempt at a solution. You actually have it.
 
Do not bother with expanding with (x+6) as well:
\frac{\sqrt{x+3}-3}{(x-6)}=\frac{x+3-9}{(x-6)(\sqrt{x+3}+3)}=\frac{(x-6)}{(x-6)(\sqrt{x+3}+3)}
 
"Look more carefully at your attempt at a solution. You actually have it."
I don't see.
\lim_{x\to 6}\frac{\sqrt{x+3}-3}{x-6}=\lim_{x\to 6}\frac{(x+3-9)(x+6)}{(x^2-36)(\sqrt{x+3}+3)}=\lim_{x\to 6}\frac{x^2+6x-6x-36}{x^2\sqrt{x+3}+3x^2-36\sqrt{x+3}-108}

I get it.
\lim_{x\to 6}\frac{\sqrt{x+3}-3}{x-6}=\lim_{x\to 6}\frac{x+3-9}{(x-6)(\sqrt{x+3}+3)}=\lim_{x\to 6}\frac{1}{\sqrt{x+3}+3}=\frac{1}{6}
 
Last edited:
How can you write x^2-36?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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