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

Limits, differentiation, related rates, integration, trig integrals, etc.

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

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!
halocreamdevil

Posts: 1
Joined: Mon May 18, 2009 3:01 am

### Re: Finding a limit via conjugates

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 ...

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

Martingale

Posts: 363
Joined: Mon Mar 30, 2009 1:30 pm
Location: USA

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...?

stapel_eliz

Posts: 1797
Joined: Mon Dec 08, 2008 4:22 pm