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by **halocreamdevil** on Mon May 18, 2009 3:10 am

Find the limit of 4 when the square root of x + 5-3 is divided by x-4.

I'm really not sure what to do. I get to the point where I've given up on everything but a table, and multiplying the bottom by the conjugate does not help because it only makes the top more complex. Please help!

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by **Martingale** on Mon May 18, 2009 6:09 am

halocreamdevil wrote:Find the limit of 4 when the square root of x + 5-3 is divided by x-4.

I'm really not sure what to do. I get to the point where I've given up on everything but a table, and multiplying the bottom by the conjugate does not help because it only makes the top more complex. Please help!

$\lim_{x\to4}\frac{\sqrt{x+5}-3}{x-4}$

if this is the problem then ...

$\lim_{x\to4}\frac{\sqrt{x+5}-3}{x-4}=\lim_{x\to4}\frac{\sqrt{x+5}-3}{x-4}\cdot\frac{\sqrt{x+5}+3}{\sqrt{x+5}+3}=\lim_{x\to4}\frac{x+5-9}{(x-4)(\sqrt{x+5}+3)}=\cdots$

by **stapel_eliz** on Mon May 18, 2009 12:14 pm

halocreamdevil wrote:...multiplying the bottom by the conjugate does not help because it only makes the top more complex.

Since multiplying the by conjugate of the radical numerator is intended to simplify "the top", it might help if you showed what you'd done...? :confused:

The Purplemath Forums

Finding limit via conjugates: sqrt x + 5 - 3 div. by x - 4

Finding limit via conjugates: sqrt x + 5 - 3 div. by x - 4

Re: Finding a limit via conjugates