# Thread: I'm not sure if I got this one right

1. ## I'm not sure if I got this one right

$\displaystyle \lim_{x\to 0} = \frac{\sqrt{x+5}-\sqrt{5}}{x}$

and what I got was

$\displaystyle \frac{1}{\sqrt{5}+\sqrt{5}}$

2. Originally Posted by drkidd22
$\displaystyle \lim_{x\to 0} = \frac{\sqrt{x+5}-\sqrt{5}}{x}$
$\displaystyle \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {x + 5} - \sqrt 5 }} {x} = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {\sqrt {x + 5} - \sqrt 5 } \right)\left( {\sqrt {x + 5} + \sqrt 5 } \right)}} {{x\left( {\sqrt {x + 5} + \sqrt 5 } \right)}} =$

$\displaystyle = \mathop {\lim }\limits_{x \to 0} \frac{{x + 5 - 5}} {{x\left( {\sqrt {x + 5} + \sqrt 5 } \right)}} = \mathop {\lim }\limits_{x \to 0} \frac{1}{{\sqrt {x + 5} + \sqrt 5 }} =$

$\displaystyle = \frac{1}{{\sqrt {0 + 5} + \sqrt 5 }} = \frac{1}{{2\sqrt 5 }} = \frac{{\sqrt 5 }}{{10}}.$

3. Oh ok, so I did do it right, just didn't complete the whole thing.

4. Originally Posted by drkidd22
$\displaystyle \lim_{x\to 0} = \frac{\sqrt{x+5}-\sqrt{5}}{x}$

and what I got was

$\displaystyle \frac{1}{\sqrt{5}+\sqrt{5}}$
In the future, when you ask for a check on what you did, it would be better to show how you got the result rather than just the result.
Rationalize the numerator by multiplying numerator and denominator by $\displaystyle \sqrt{x+5}+ \sqrt{5}$:
$\displaystyle \frac{x+ 5- x}{x(\sqrt{x+5}+ \sqrt{x})}= \frac{x}{x(\sqrt{x+5}+ \sqrt{5}}$$\displaystyle \frac{1}{\sqrt{x+5}+ \sqrt{5}}$.
Now you can put x= 0 and get $\displaystyle \frac{1}{\sqrt{5}+\sqrt{5}}$, what you have. But I would recommend that you write as $\displaystyle \frac{1}{2\sqrt{5}}$ or even $\displaystyle \frac{\sqrt{5}}{10}$.

Blast! DeMath got in ahead of me!

5. This is anotherone I tried doing today.

$\displaystyle \lim_{x\to 0} = \frac{\frac{1}{3+x}-\frac{1}{3}}{x}$

$\displaystyle \frac{\frac{3-3+x}{3(3+x)}}{x}$

$\displaystyle \frac{\frac{x}{9+3x}}{x} =9?$

6. Originally Posted by drkidd22
This is anotherone I tried doing today.

$\displaystyle \lim_{x\to 0} = \frac{\frac{1}{3+x}-\frac{1}{3}}{x}$

$\displaystyle \frac{\frac{3-3+x}{3(3+x)}}{x}$

$\displaystyle \frac{\frac{x}{9+3x}}{x} =9?$
$\displaystyle \frac{\frac{3-(3+x)}{3(3+x)}}{x} = \frac{-x}{3x(3+x)} = \frac{-1}{3(3+x)}$

limit is $\displaystyle -\frac{1}{9}$

in future, start a new problem with a new thread.