1. ## Limits

Evaluate lim x->0 [√(1+x) -1]/x

when I sub in 0 i get 0/0

I am trying to simply the expression or cancel terms but I am having no luck. the answer at the back of the book is 1/2.

2. ## Re: Limits

Originally Posted by edwardkiely
Evaluate lim x->0 [√(1+x) -1]/x

when I sub in 0 i get 0/0

I am trying to simply the expression or cancel terms but I am having no luck. the answer at the back of the book is 1/2.
Rationalize the numerator ...

$\dfrac{\sqrt{1+x}-1}{x} \cdot \dfrac{\sqrt{1+x}+1}{\sqrt{1+x}+1}$

... then determine the limit.

3. ## Re: Limits

when I sub 0 into the simplified version I get 0

4. ## Re: Limits

Originally Posted by skeeter
Rationalize the numerator ...

$\dfrac{\sqrt{1+x}-1}{x} \cdot \dfrac{\sqrt{1+x}+1}{\sqrt{1+x}+1}$

... then determine the limit.
$\dfrac{\sqrt{1+x}-1}{x} \cdot \dfrac{\sqrt{1+x}+1}{\sqrt{1+x}+1} = \dfrac{(1+x)-1}{x(\sqrt{1+x}+1} = \dfrac{1}{\sqrt{1+x}+1}$

$\lim \limits_{x\to 0} \dfrac{1}{\sqrt{1+x}+1} = \dfrac{1}{\sqrt{1+0}+1} = \dfrac 1 2$

5. ## Re: Limits

$\dfrac{\sqrt{1+x}-1}{x} \cdot \dfrac{\sqrt{1+x}+1}{\sqrt{1+x}+1} = \frac{ (1+x)-1}{x\left(\sqrt{1+x}+1\right)} = \frac {x}{x\left(\sqrt{1+x}+1\right)} = \frac {1}{\sqrt{1+x}+1}$