Limit of (2-sqrt(x))/(3-sqrt(2x+1)) as x approaches 4 - Simple Limits Question

[Yalladoonia]

Homework Statement

Evaluate the following limit
[itex]
\displaystyle\lim_{x\rightarrow 4} {\frac{2-\sqrt{x}}{3-\sqrt{2x+1}}}[/itex]

Homework Equations

lim x -> 4

The Attempt at a Solution

i tried change of variables so i get [itex]√2x+1 = u[/itex]
then i rearranged that to get [itex]u^2-1/2[/itex]
And i rearrnage original equation and i get this..
[itex]-(u^2-7)/3-u(u^2-1/2)[/itex]

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

 

[Yalladoonia]

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

 

[dirk_mec1]

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

 

[BiGyElLoWhAt]

Or l'hopital's?

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

##\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

 

[Yalladoonia]

i think l'hopitals is derivatives, i haven't learned that yet I am only allowed to use limits

 

[Mark44]

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.