Calculating a basic limit

  • Thread starter gipc
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  • #1
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Hello,

can someone please help me understand how to solve the following limit? I've tried multiplying by sqrt(x)+sqrt(a) but it doesn't seem to do the trick. How do i continue from there?

http://img78.imageshack.us/img78/3595/asdsadassssggg.jpg [Broken]
 
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Answers and Replies

  • #2
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L'Hôpital's rule does the job here, check http://en.wikipedia.org/wiki/L'Hôpital's_rule , but as I'm not such a fan of L'Hôpital there is usually a way to work yourself around hopital but I don't see it at this moment. Again, use L'Hôpital for an easy way out here!
 
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  • #3
34,885
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Hello,

can someone please help me understand how to solve the following limit? I've tried multiplying by sqrt(x)+sqrt(a) but it doesn't seem to do the trick. How do i continue from there?

http://img78.imageshack.us/img78/3595/asdsadassssggg.jpg [Broken]
The[/URL] two-sided limit doesn't exist, because if x < a, then the denominator is not a real number. The right-side limit exists, though. If you assume that x > a, multiplying numerator and denominator by sqrt(x) + sqrt(a) will get you something that you can evaluate.
 
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  • #4
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Ah, took me a while but you should rather multiply nominator and denominator [tex](x^{\frac{1}{2}}+a^{\frac{1}{2}})(x+a)[/tex] so you can REALLY evaluate the limit with ease :)
 
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  • #5
34,885
6,624
I repeat - the two-sided limit doesn't exist, so if you get a value for it, your work is wrong.
 
  • #6
34,885
6,624
Ah, took me a while but you should rather multiply nominator and denominator [tex]x^{\frac{3}{2}}+a^{\frac{3}{2}}[/tex] so you can REALLY evaluate the limit with ease :)
And how does that work? Are you saying that (x1/2 - a1/2)(x3/2 + a3/2) gives you something easy to work with? The middle terms do not drop out.
 
  • #7
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You are right, sorry, I meant multiplying by (x+a)(sqrt(x)+sqrt(a)) but didn't think it over.
 

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