# One sided Limits

• Jan 13th 2009, 04:13 PM
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
$(1-2x)/(x^2-1)$

Could someone provide me a precise way to show the limit goes to infinity?
• Jan 13th 2009, 04:59 PM
chipai
I use a theorem : let ${x_k}->1$ from left as k->infinite , and f be continuous if ${f(x_k)}->infinite$ , $limf(x)=infinite$as $x->1^-$

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

And $f(x_k)={(k^2)/2}-k/(4k-2)$ this is infinite as k->infinite
• Jan 13th 2009, 05:14 PM
Quote:

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

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

And $f(x_k)={(k^2)/2}-k/(4k-2)$ 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 ${x_k}={(k-1)/k}$. It is an increasing function, but is there a way to find such {x_k}? EDIT: or is it arbitrary?
• Jan 13th 2009, 05:22 PM
chipai
Since a divergent sequence can has a convergent subsequence ,the sequence i used is not arbitary.
• Jan 13th 2009, 05:25 PM
Quote:

Originally Posted by chipai
Since a divergent sequence can has a convergent subsequence ,the sequence i used is not arbitary.

thank you for your help. :)
• Jan 13th 2009, 05:30 PM
NonCommAlg
think simple: $x^2 - 1$ approaches 0 from negative values because $x \rightarrow 1-,$ and $1 - 2x$ approaches $-1.$ so the limit is $+\infty.$
• Jan 13th 2009, 05:51 PM
think simple: $x^2 - 1$ approaches 0 from negative values because $x \rightarrow 1-,$ and $1 - 2x$ approaches $-1.$ so the limit is $+\infty.$