# Seeking help for limit math problem

1. Sep 19, 2007

### teffy3001

lim (sqrt(x^2+5) - 3)/(x-2)
x->2

i hope that kinda makes sense...

anyways, my question is that ive been doing this and i keep getting 0 in the denominator in the answer...after subsituting everything i got 4/ sqrt(9) - 3 = 4/0
so then is this equation just not possible?

Last edited: Sep 19, 2007
2. Sep 19, 2007

### matt grime

How did you get this? That doesn't appear to bear any relation to the initial expression you gave.

3. Sep 19, 2007

For $$\lim_{x \rightarrow 2}\frac{\sqrt{x^2+5}-3}{x-2}$$

Clearly the substitution method will give you a zero denominator!

So, what methods have you learned to try and "avoid" a zero denominator?

Casey

I think they meant (sqrt(4+5)-3)/0....
Arithmetic errors are always a good start

Last edited: Sep 19, 2007
4. Sep 19, 2007

### teffy3001

um the course im taking told me first multiply the numerator and denominator by
sqrt(x^2+5) - 3

i might have done it wrong but then i got x^2+5-9/x-2(sqrt(x^2+5) - 3)
i factored that to x^2-4/x-2(sqrt(x^2+5) - 3)
and then eventually got (x-2)(x+2)/x-2(sqrt(x^2+5) - 3)
canceled out x-2
x+2/sqrt(x^2 + 5) - 3
then subtituted the x for 2
thats how i got that answer....i think thats how the teacher told me to do it...probably made mistakes along the way though..im not sure

5. Sep 19, 2007

### EnumaElish

Where is the interaction term in your numerator? How do you expand (a + b)^2?

6. Sep 19, 2007

the conjugate of $$\sqrt{x^2+5}-3$$ is $$\sqrt{x^2+5}+3$$
Hint: $$(a+b)^2$$ does not = $$a^2+b^2$$