1. ## Basic Limits

Hi, I need a hand evaluating a couple of limits without using L'hopital's.
I've done a ton of these so far and I often need to use little 'tricks' like factorising/cancelling etc, but I can't see these two.

$\frac{\sqrt{x+2}-3}{x-7}$ as x approaches 7

and

$\frac{\frac{1}{4} + \frac{1}{x}}{4+x}$ as x approaches -4.

Cheers.

2. Originally Posted by U-God
Hi, I need a hand evaluating a couple of limits without using L'hopital's.
I've done a ton of these so far and I often need to use little 'tricks' like factorising/cancelling etc, but I can't see these two.

$\frac{\sqrt{x+2}-3}{x-7}$ as x approaches 7
Multiply by the conjugate of the numerator:

$\lim_{x\to7}\frac{\sqrt{x+2}-3}{x-7}=\lim_{x\to 7}\frac{\sqrt{x+2}-3}{x-7}\cdot\frac{\sqrt{x+2}+3}{\sqrt{x+2}+3}=\lim_{x\t o7}\frac{x+2-9}{(x-7)(\sqrt{x+2}+3)}$

Can you take it from here?

and

$\frac{\frac{1}{4} + \frac{1}{x}}{4+x}$ as x approaches -4.

Cheers.
Combine the terms in the numerator. The LCD is 4x:

$\lim_{x\to -4}\frac{\frac{x}{4x} + \frac{4}{4x}}{4+x}=\lim_{x\to-4} \frac{\frac{4+x}{4x}}{4+x}$

Can you take it from here?

--Chris

3. Originally Posted by Chris L T521
Multiply by the conjugate of the numerator:

$\lim_{x\to7}\frac{\sqrt{x+2}-3}{x-7}=\lim_{x\to 7}\frac{\sqrt{x+2}-3}{x-7}\cdot\frac{\sqrt{x+2}+3}{\sqrt{x+2}+3}=\lim_{x\t o7}\frac{x+2-9}{(x-7)(\sqrt{x+2}+3)}$

Can you take it from here?

Combine the terms in the numerator. The LCD is 4x:

$\lim_{x\to -4}\frac{\frac{x}{4x} + \frac{4}{4x}}{4+x}=\lim_{x\to-4} \frac{\frac{4+x}{4x}}{4+x}$

Can you take it from here?

--Chris
I certainly can take it from there, thanks tons

4. Originally Posted by U-God
Hi, I need a hand evaluating a couple of limits without using L'hopital's.
I've done a ton of these so far and I often need to use little 'tricks' like factorising/cancelling etc, but I can't see these two.

$\frac{\sqrt{x+2}-3}{x-7}$ as x approaches 7

and

$\frac{\frac{1}{4} + \frac{1}{x}}{4+x}$ as x approaches -4.

Cheers.
For the first, Take the derivative of the top and the derivative of the bottom, then sub 7 into both.

5. Originally Posted by Prove It
For the first, Take the derivative of the top and the derivative of the bottom, then sub 7 into both.
Thanks, but I can't use l'hopital's rule