Dobin Posted February 4, 2007 Report Share Posted February 4, 2007 Just reviewing a bit of calculus, and this question stumped me. [Latex]\displaystyle\lim_{x\rightarrow2} \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1} [/math]Anyone got a solution? And i would appreciate an algebraic way, you dont learn anything from just typing it into your calculator. Edit: Ive already tried rationalizing the top and bottom, neither one creates an equation in which you can derive an answer. Thanks. PS. that latex stuff can be confusing :xx: Quote Link to comment Share on other sites More sharing options...

ronthepon Posted February 4, 2007 Report Share Posted February 4, 2007 [math]\displaystyle\lim_{x\rightarrow2} \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1} [/math] This is a pretty simple question, once you figure the basics behind it. [math]= \displaystyle\lim_{x\rightarrow2} \frac{(6-x-4)(\sqrt{3-x}+1)}{(3-x-1)(\sqrt{6-x}+2)} [/math] See? I've rationalised it. Pretty simple now. [math]= \displaystyle\lim_{x\rightarrow2} \frac{(2-x)(\sqrt{3-x}+1)}{(2-x)(\sqrt{6-x}+2)} [/math] What do we see here? 2-x. Remember that whenever an algebric term tends to zero for a value of it's variable, then you always have one or more factors that become zero at that value. All you need to do is manipulate the expression in such a way so as to 'expose' that factor. I'd say that the things are pretty simple from here. Go on, and feel free to ask if you don't figure. Quote Link to comment Share on other sites More sharing options...

Dobin Posted February 4, 2007 Author Report Share Posted February 4, 2007 Well. I'm surprised I didnt see that. Thanks Ron. Quote Link to comment Share on other sites More sharing options...

LJP07 Posted February 4, 2007 Report Share Posted February 4, 2007 That is not the way I would have done it. :hihi: Normally you would multiply above and below by the conjugate surd. So you would have the above starting equation multiplied above and below by root 3-x plus one. Quote Link to comment Share on other sites More sharing options...

ronthepon Posted February 4, 2007 Report Share Posted February 4, 2007 That's what we've done, except that I've simultaneously rationalised the numerator as well. (Take a closer look at the brackets) Quote Link to comment Share on other sites More sharing options...

Nootropic Posted February 16, 2007 Report Share Posted February 16, 2007 Is it just me or would it even be necessary to rationalize anything? Direct substiution, in this case, is all that's necessary, since it ends up being the square root of two over 0, indicating the fact that a limit doesn't exist there. Or is there something I'm missing? Quote Link to comment Share on other sites More sharing options...

Dobin Posted February 16, 2007 Author Report Share Posted February 16, 2007 Nootropic, if you use direct substitution, you get 0/0, which is undefined, not the root of 2 over 0. Quote Link to comment Share on other sites More sharing options...

Nootropic Posted February 16, 2007 Report Share Posted February 16, 2007 oh, haha, whoops, guess I can't see the extension of the square root sign. Quote Link to comment Share on other sites More sharing options...

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