# Having problem solving a limit

1. Oct 13, 2011

### the_morbidus

having problem solving a limit!!

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

Solve the following limits algebraically.

lim x->5 x^2 - 25 / √(2x+6) -4

2. Relevant equations

3. The attempt at a solution

i've tried a different few ways.

so by factoring the top i have (x+5)(x-5) so i get x=5 or x=-5, and while 5 is the limit, -5 on top with the square it will just render the negative sign useless.

tried to multiply the bottom cognitive
(x^2-25)(√(2x+6)+4)/(√(2x+6)-4)(√(2x+6)+4) =
=(x^2-25)(√(2x+6)+4)/(√(2x+6))^2 - (4)^2 =
=(x^2-25)(√(2x+6)+4)/2x+6-16=
=(x^2-25)(√(2x+6)+4)/2x-10

i'm stuck here and i doubt this is even the proper route.

2. Oct 13, 2011

### Staff: Mentor

Re: having problem solving a limit!!

Multiplying by the conjugate over itself is a useful approach.

$$\lim_{x \to 5}\frac{(x-5)(x+5)}{\sqrt{2x+6}-4}\cdot \frac{\sqrt{2x+6} + 4}{\sqrt{2x+6} + 4}$$
$$= \lim_{x \to 5}\frac{(x-5)(x+5)(\sqrt{2x+6} + 4)}{(\sqrt{2x+6}-4) \cdot (\sqrt{2x+6} + 4)}$$
Can you take it from here?

3. Oct 13, 2011

### symbolipoint

Re: having problem solving a limit!!

What you really mean is
$$Lim_{x\rightarrow 5} \frac{x^2-25}{\sqrt{2x+6}-4}$$

...

Most of your work looks good. Try then dividing numerator and denominator by either x, or x^2 ?

4. Oct 13, 2011

### symbolipoint

Re: having problem solving a limit!!

He appeared to be doing that; maybe I was confused looking at all the steps and symbols through text?
EDIT: Give me the chance to try the problem myself; maybe more comment later.

OK, very neat. Just multiplying numerator and denominator by conjugate gives you something to simplify and then simple substitution of x=5 can be evaluated with no complications.

Last edited by a moderator: Oct 13, 2011
5. Oct 13, 2011

### HallsofIvy

Staff Emeritus
Re: having problem solving a limit!!

Both numerator and denominator are 0 when x= 5 therefore both have a factor of x- 5:
$$\frac{(x- 5)(x+ 5)(\sqrt{2x+ 6}+ 4)}{2(x- 5)}$$

6. Oct 13, 2011

### the_morbidus

Re: having problem solving a limit!!

thank you guys, when I woke up this morning and having looked at a few other problems I arrived at the same solution as HallsofIvy, I then removed the (x-5) and substituted the x for 5 and i got 40 as the limit I think.

7. Oct 13, 2011

### symbolipoint

Re: having problem solving a limit!!

Yes, and in fact, you both think and KNOW for sure now.