Can someone tell me the limit of (x/(2x-2))-(1/((x^2)-1)) as x approaches 1.

Can someone tell me the limit of (x/(2x-2))-(1/((x^2)-1)) as x approaches 1.

Rewrite the fractions, expand the second one, add them and simplify.

 

[math]

\begin{gathered}

\frac{x}

{{2x - 2}} - \frac{1}

{{x^2 - 1}} \hfill \\

\frac{x}

{{2\left( {x - 1} \right)}} - \frac{1}

{{\left( {x - 1} \right)\left( {x + 1} \right)}} \hfill \\

\frac{x}

{{2\left( {x - 1} \right)}} - \left( {\frac{1}

{{2\left( {x - 1} \right)}} - \frac{1}

{{2\left( {x + 1} \right)}}} \right) \hfill \\

\frac{{x + 2}}

{{2\left( {x + 1} \right)}} \hfill \\

\end{gathered}

[/math]

 

Now fill in x = 1

Thank you so much!

No problem

[math]\frac{x}{2(x-1)}-\frac{1}{(x-1)(x+1)}[/math]

 

[math]\frac{x}{2(x-1)}-(\frac{1}{2(x-1)}-\frac{1}{2(x+1)})[/math]

 

How do you figure that?

Haven't checked it, but probably partial fractions.

Yes, you can split using partial fractions or by "playing with fractions" like this:

 

Yes' date=' you can split using partial fractions or by "playing with fractions" like this:

 

[img']http://ld.livedelivery.com/F/2157642/Image.gif[/img]

 

I don't see how any of the second half of these work. It seems to me to not work. Is this something I'd learn after Calculus I?

Well normally you'd split by doing partial fractions but what I did doesn't require any advanced calculus. I added and substracted same things in the nominator (e.g. replaced "1" by "1+x-x" in the first step) and then split the fraction to simplify etc...