Solving a Basic Limit Problem Using Conjugate Multiplication

[Dissonance in E]

Homework Statement

Find the limit of $$(x-3)/(\sqrt{1+x}-2)$$ as x tends to 3

Homework Equations

Conjugate multiplication.

The Attempt at a Solution

$$(x-3)(\sqrt{1+x}+2)/(\sqrt{1+x}-2)(\sqrt{1+x}+2)$$

$$(x\sqrt{1+x}+2x-3\sqrt{1+x}-6)/x-3$$

This is where i get stuck, I am thinking to get rid of the x in the denominator but the -6 in the numerator is what stumps me. Am i supposed to just factor the top somehow?

 

[LCKurtz]

Don't multiply the numerator out. You have a factor of x - 3 to work with.

 

[Mark44]

When you multiply by 1 in the form of (sqrt(1 + x) + 2) over itself, you should get
$$\frac{(x - 3)(\sqrt{1 + x} + 2}{x - 3}$$

I think you made a mistake in multiplying your original denominator by its conjugate.

 

[Dissonance in E]

Oh man, how could I not see that. Thanks!