I'm solving a limit problem that looks something like this:

lim as x-> 0 ; 1-(sqrt) 2x(sq)-1 / x-1

each time I attempt to solve this problem, I end up solving in circles and I'm uncertain what I should do. We've only learned basic things about limits and hence I'm confused as to where to start besides multiplying by the 1+ (sqrt) 2x(sq)-1 / 1+ (sqrt)2x(sq)-1. after I've done that I get something to the effect of 1-2x(sq)-1/ (x-1) (1+(sqrt)2x(sq)-1);; can someone please tell me if my math is flawed or what I should do next?

Originally Posted by BernieBriggs
I'm solving a limit problem that looks something like this:
lim as x-> 0 ; 1-(sqrt) 2x(sq)-1 / x-1
It the problem is $\displaystyle \lim _{x \to 0} \frac{{1 - \sqrt {2x^2 - 1} }}{{x - 1}}$, then the limit does not exist, because 0 is not in the domain of the function.

rather than try "fancy algebraic manipulation" just plug in x = 0. do you notice something odd about the square root?

It comes out as sqrt of -1; which i know is imaginary. the problem is that my math teacher insists that you can solve it but I don't even understand how.;; and from what you both said it doesn't exist.

****EDITTTT!;*** he changed the problem to x approaches 1; what does this change for this problem

Originally Posted by BernieBriggs
****EDITTTT!;*** he changed the problem to x approaches 1; what does this change for this problem
In that case:
$\displaystyle \frac{{1 - \sqrt {2x^2 - 1} }}{{x - 1}}=\frac{-2(x+1)(x-1)}{(x-1)(1 + \sqrt {2x^2 - 1})}$

It changes it crucially! x= 1 is in at least on the boundary of the domain of this function. Of course if you just put x= 1 in the formula you get "0/0" which does not give an answer. But it should tell you that you need to do some factoring. looking at that square root in the numerator suggests to me that you should "rationalize the numerator".

I see what I did wrong; I was just factoring in correctly. Thank you all so much!

ps; just to affirm I'm not crazy; x(sq)-2x+4; if you plug in -2; do you get 12? (unrelated problem; but someone tried to explain this equals 10 and I was like ? no...)

alright; so now i've factored correctly but I'm still caught in this horrible paradoxic multiplication of roots. i divided out and then multiplied out the root to the top; so now i have 2(x+1) (1- sqrt 2x(sq)-1)/2-2x(sq).

what've I done wrong, now? : / I'm still getting zero as a denominator and I'm not sure how to fix it.

$\displaystyle \lim _{x \to 1} \frac{{1 - \sqrt {2x^2 - 1} }}{{x - 1}} \cdot \frac{1 + \sqrt {2x^2 - 1}}{1 + \sqrt {2x^2 - 1}}$

$\displaystyle \lim _{x \to 1} \frac{1 - (2x^2 - 1)}{(x-1)(1 + \sqrt {2x^2 - 1})}$

$\displaystyle \lim _{x \to 1} \frac{2 - 2x^2}{(x-1)(1 + \sqrt {2x^2 - 1})}$

$\displaystyle \lim _{x \to 1} \frac{2(1-x^2)}{(x-1)(1 + \sqrt {2x^2 - 1})}$

$\displaystyle \lim _{x \to 1} \frac{2(1-x)(1+x)}{(x-1)(1 + \sqrt {2x^2 - 1})}$

$\displaystyle \lim _{x \to 1} \frac{-2(1+x)}{(1 + \sqrt {2x^2 - 1})}$

now perform the direct substitution of $\displaystyle x = 1$