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?

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- Jan 13th 2009, 04:13 PMMacstersUndeadOne 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? - Jan 13th 2009, 04:59 PMchipai
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 - Jan 13th 2009, 05:14 PMMacstersUndead
- Jan 13th 2009, 05:22 PMchipai
Since a divergent sequence can has a convergent subsequence ,the sequence i used is not arbitary.

- Jan 13th 2009, 05:25 PMMacstersUndead
- Jan 13th 2009, 05:30 PMNonCommAlg
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 $\displaystyle +\infty.$ - Jan 13th 2009, 05:51 PMMacstersUndead