1. ## One sided Limits

I was posed this problem on another forum by a first year and for some reason I'm struggling

lim x-> 1 from the left of
$\displaystyle (1-2x)/(x^2-1)$

Could someone provide me a precise way to show the limit goes to infinity?

2. I use a theorem : let $\displaystyle {x_k}->1$ from left as k->infinite , and f be continuous if $\displaystyle {f(x_k)}->infinite$ , $\displaystyle limf(x)=infinite$as $\displaystyle x->1^-$

clearly , $\displaystyle (1-2x)/(x^2-1)$ is a continuous on $\displaystyle (-infinite , 1^-)$ and let $\displaystyle {x_k}={(k-1)/k}$converges to 1 from left.

And $\displaystyle f(x_k)={(k^2)/2}-k/(4k-2)$ this is infinite as k->infinite

3. Originally Posted by chipai
I use a theorem : let $\displaystyle {x_k}->1$ from left as k->infinite , and f be continuous if $\displaystyle {f(x_k)}->infinite$ , $\displaystyle limf(x)=infinite$as $\displaystyle x->1^-$

clearly , $\displaystyle (1-2x)/(x^2-1)$ is a continuous on $\displaystyle (-infinite , 1^-)$ and let $\displaystyle {x_k}={(k-1)/k}$converges to 1 from left.

And $\displaystyle f(x_k)={(k^2)/2}-k/(4k-2)$ this is infinite as k->infinite
Oh yes! I have heard a theorem like that before. So I follow you up until the point where you let $\displaystyle {x_k}={(k-1)/k}$. It is an increasing function, but is there a way to find such {x_k}? EDIT: or is it arbitrary?

4. Since a divergent sequence can has a convergent subsequence ,the sequence i used is not arbitary.

5. Originally Posted by chipai
Since a divergent sequence can has a convergent subsequence ,the sequence i used is not arbitary.
6. think simple: $\displaystyle x^2 - 1$ approaches 0 from negative values because $\displaystyle x \rightarrow 1-,$ and $\displaystyle 1 - 2x$ approaches $\displaystyle -1.$ so the limit is $\displaystyle +\infty.$
think simple: $\displaystyle x^2 - 1$ approaches 0 from negative values because $\displaystyle x \rightarrow 1-,$ and $\displaystyle 1 - 2x$ approaches $\displaystyle -1.$ so the limit is $\displaystyle +\infty.$