# Homework Help: Simple Limits Question

1. Oct 6, 2014

1. The problem statement, all variables and given/known data

Evaluate the following limit
$\displaystyle\lim_{x\rightarrow 4} {\frac{2-\sqrt{x}}{3-\sqrt{2x+1}}}$
2. Relevant equations

lim x -> 4
3. The attempt at a solution
i tried change of variables so i get $√2x+1 = u$
then i rearranged that to get $u^2-1/2$
And i rearrnage original equation and i get this..
$-(u^2-7)/3-u(u^2-1/2)$

Sorry i am new with this latex thing .. i don treally konw how to use it the sqrt thing so i just lay it out this way

Last edited by a moderator: Oct 6, 2014
2. Oct 6, 2014

I am really unsure of how to use this latex thing... but personally i think my attempt at the solution is all garbage so if anyone could guide me on the right path that would be help ful

3. Oct 6, 2014

### dirk_mec1

Multiply with the nominator and denominator with 3+sqrt(4x+1)

4. Oct 6, 2014

### BiGyElLoWhAt

Or l'hopital's?
$\frac{2 - \sqrt{x}}{3 - \sqrt{2x + 1}}\frac{3+\sqrt{2x+1}}{3+\sqrt{2x+1}}=\frac{(2-\sqrt{x})(3+\sqrt{2x+1})}{9 - (2x +1)}$
which still has a divide by zero error as x->4

5. Oct 6, 2014

i think l'hopitals is derivatives, i havent learned that yet im only allowed to use limits

6. Oct 6, 2014

### Staff: Mentor

You need to rationalize both the numerator and denominator. To do this, multiply by the following:
$$\frac{2 + \sqrt{x}}{2 + \sqrt{x}} \cdot \frac{3 + \sqrt{2x + 1}}{3 + \sqrt{2x + 1}}$$

When you do this, you get something that you can simplify and then take the limit.