# Thread: Help. Can I do this? (Delta Epsilon)

1. ## Help. Can I do this? (Delta Epsilon)

Hi everyone. I'm a bit stuck on doing this Delta Epsilon prrof for the limit
lim x->1 $\displaystyle \sqrt{x}$=1
I have (I'll use brakets for absolute values)
( $\displaystyle \sqrt{x}$-1)<e and 0<(x-1)<d
so
(x-1)<$\displaystyle e^s$
so d=$\displaystyle e^2$ ?????

ok then i'm stuck and don't know where from here...i'm not even sure if i have done that right

any help?

2. Can you show that if $\displaystyle |x-1|<1$ then $\displaystyle \frac{1}{{\sqrt x + 1}} < 1?$
If you can then choose $\displaystyle \delta = \min \left\{ {1,\varepsilon } \right\}$.

3. Hi
If i have f(x)-L < e
then sqrt(x)-1 < e
if i then multiply by ($\displaystyle \sqrt{x}+1$)
i get x-1 < e($\displaystyle \sqrt{x}+1$)
is that right?

but if [x-1] = ($\displaystyle \sqrt{x}-1$)($\displaystyle \sqrt{x}+1$)

if i then cancel from both sides i'm back where i started at $\displaystyle \sqrt{x}-1$<e

I think i just keep turning back on myself....I'm not being able to get a 'd'