1. ## limits help

problem:

lim as x -> 4 : 4-x / (5 - sqrt(x^2 + 9))

I tried to use the conjugate property but in the end when I substitute 4 into the equation I come up with it turns to:

(4 - 4) * (5 + sqrt(4^2 + 9)) / 25 - 16 + 9
0 * 5 + 5 / 18

the answer should be 5/4, where am I going wrong with this problem?

2. ## Re: limits help

Originally Posted by deltemis
problem:

lim as x -> 4 : 4-x / 5 - sqrt(x^2 + 9)

I tried to use the conjugate property but in the end when I substitute 4 into the equation I come up with it turns to:

(4 - 4) * (5 + sqrt(4^2 + 9)) / 25 - 16 + 9
0 * 5 + 5 / 18

the answer should be 5/4, where am I going wrong with this problem?
we need to get the parens squared away first. Is this what you mean?

$\displaystyle{\lim_{x \to 4}}\frac{4-x}{5}-\sqrt{x^2+9}$

or do you maybe mean this?

$\displaystyle{\lim_{x \to 4}}\frac{4-x}{5-\sqrt{x^2+9}}$

I think you mean the 2nd one.

$\frac{4-x}{5-\sqrt{x^2+9}}=\frac{4-x}{5-\sqrt{x^2+9}}\frac{5+\sqrt{x^2+9}}{5+\sqrt{x^2+9}} =$

$\frac{(4-x)\left(5+\sqrt{x^2+9}\right)}{5^2-(x^2+9)}=\frac{(4-x)\left(5+\sqrt{x^2+9}\right)}{16-x^2}=$

$\frac{(4-x)\left(5+\sqrt{x^2+9}\right)}{(4-x)(4+x)}=\frac{5+\sqrt{x^2+9}}{4+x}$

Now you can just plug 4 in for x to find the limit

$\displaystyle{\lim_{x \to 4}}\frac{4-x}{5+\sqrt{x^2+9}}=\frac{5+\sqrt{4^2+9}}{(4+4)}=$

$\frac{10}{8}= \frac{5}{4}$

3. ## Re: limits help

it's what you have but the sqrt portion is part of the denominator

4. ## Re: limits help

Originally Posted by deltemis
problem:

lim as x -> 4 : 4-x / 5 - sqrt(x^2 + 9)

I tried to use the conjugate property but in the end when I substitute 4 into the equation I come up with it turns to:
(4 - 4) * (5 + sqrt(4^2 + 9)) / 25 - 16 + 9
0 * 5 + 5 / 18

the answer should be 5/4, where am I going wrong with this problem?
I think that you mean something else altogether.
$\frac{4-x}{5-\sqrt{x^2+9}}=\frac{(4-x)(5+\sqrt{x^2+9})}{25-(x^2+9)}=\frac{5+\sqrt{x^2+9}}{4+x}$