Rationalising the denominator Limit question Watch

username970964
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#1
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So, the question is in the first line. This is my working, but apparently the answer is 1/sqrt2. Where did I go wrong and what did I do wrong? I suspect it's something from the 3rd to the 4th line. Am I suppose to rationalise the denominator at all? Or is it a different method?Name:  math1.jpg
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Smaug123
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(Original post by Airess3)
So, the question is in the first line. This is my working, but apparently the answer is 1/sqrt2. Where did I go wrong and what did I do wrong? I suspect it's something from the 3rd to the 4th line. Am I suppose to rationalise the denominator at all? Or is it a different method?
The step where the limit sign vanishes is wrong: \sqrt{2-t} \times \sqrt{2+t} \not = 2-t. (Limit sign should have remained throughout, but it's a good way to identify the line.)

Do you know L'Hôpital's rule?
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TeeEm
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(Original post by Airess3)
So, the question is in the first line. This is my working, but apparently the answer is 1/sqrt2. Where did I go wrong and what did I do wrong? I suspect it's something from the 3rd to the 4th line. Am I suppose to rationalise the denominator at all? Or is it a different method?Name:  math1.jpg
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you multiply top and bottom by the surd conjugate [2 + root(t) + (2 - root(t) ]
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TeeEm
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(Original post by Hasufel)
<- what he said!

here, i think its easier to "rationalise the numerator". you end up with, before the limit

\displaystyle \frac{2}{(\sqrt{2+t}+ \sqrt{2-t}}
???

that is exactly what I suggested
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Hasufel
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(Original post by TeeEm)
???

that is exactly what I suggested
Aherm! - my apologies. Post removed - `s not neccessary afterall!
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TeeEm
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(Original post by Hasufel)
Aherm! - my apologies. Post removed - `s not neccessary afterall!
you did not have to remove ...
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Hasufel
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(Original post by TeeEm)
you did not have to remove ...
Meh! - you explained the solution needed - `nuff said!
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username970964
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(Original post by Smaug123)
The step where the limit sign vanishes is wrong: \sqrt{2-t} \times \sqrt{2+t} \not = 2-t. (Limit sign should have remained throughout, but it's a good way to identify the line.)

Do you know L'Hôpital's rule?
No, I don't know L'Hopital's rule.
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TeeEm
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(Original post by Airess3)
No, I don't know L'Hopital's rule.
look at two similar examples (with square roots) at the very last page of this booklet

http://madasmaths.com/archive/maths_...ics/limits.pdf
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username970964
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Ok, I've made amendments but I'm still stuck on what (sqrt(2-t)) x (sqrt(2+t)) = ?
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TeeEm
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(Original post by Airess3)
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Ok, I've made amendments but I'm still stuck on what (sqrt(2-t)) x (sqrt(2+t)) = ?
Start from the beginning

Look at post 3 and maybe post 9
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username970964
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(Original post by TeeEm)
Start from the beginning

Look at post 3 and maybe post 9
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But the answer is supposed to be 1/sqrt2 but I got 0?
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TeeEm
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(Original post by Airess3)
What about now? Name:  2014-12-03 14.00.23.jpg
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But the answer is supposed to be 1/sqrt2 but I got 0?
you are applying the technique I suggested "correctly" but the multiplication which follow are all incorrect.

you must use brackets correctly
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