K9-47G Posted September 21, 2005 Share Posted September 21, 2005 Can someone tell me the limit of (x/(2x-2))-(1/((x^2)-1)) as x approaches 1. Link to comment Share on other sites More sharing options...
TD Posted September 22, 2005 Share Posted September 22, 2005 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 Link to comment Share on other sites More sharing options...
K9-47G Posted September 24, 2005 Author Share Posted September 24, 2005 Thank you so much! Link to comment Share on other sites More sharing options...
TD Posted September 24, 2005 Share Posted September 24, 2005 No problem Link to comment Share on other sites More sharing options...
BobbyJoeCool Posted September 24, 2005 Share Posted September 24, 2005 [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? Link to comment Share on other sites More sharing options...
Dave Posted September 25, 2005 Share Posted September 25, 2005 Haven't checked it, but probably partial fractions. Link to comment Share on other sites More sharing options...
TD Posted September 25, 2005 Share Posted September 25, 2005 Yes, you can split using partial fractions or by "playing with fractions" like this: Link to comment Share on other sites More sharing options...
BobbyJoeCool Posted September 26, 2005 Share Posted September 26, 2005 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? Link to comment Share on other sites More sharing options...
TD Posted September 26, 2005 Share Posted September 26, 2005 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... Link to comment Share on other sites More sharing options...
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