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Calculus help. (limits)


Dobin

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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:

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[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.

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