Solving Limits: Finding a & b When x Approaches 0

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Homework Statement



I'm doing an assignment, and I've hit a wall with this limits question:

6. If the limit as x approaches 0 of [URL]http://www2.wolframalpha.com/Calculate/MSP/MSP202919ggbg4de783c0e7000024ba5hah2cdg59e6?MSPStoreType=image/gif&s=14&w=106&h=41[/URL] equals \sqrt{5}, deduce the values of a and b.


The Attempt at a Solution



So far, I've multiplied the numerator and denominator by the conjugate, and now I have [PLAIN]http://www4d.wolframalpha.com/Calculate/MSP/MSP231719ggb7fd1fig87b800001ie7f55ge9hc9190?MSPStoreType=image/gif&s=9&w=106&h=42. I've tried squaring the bottom and the top and expanding them out, but I'm just spinning my wheels at this point, I don't really know how to proceed. I would appreciate any hints you could give me, thanks in advance.
 
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cripple said:

Homework Statement



I'm doing an assignment, and I've hit a wall with this limits question:

6. If the limit as x approaches 0 of [URL]http://www2.wolframalpha.com/Calculate/MSP/MSP202919ggbg4de783c0e7000024ba5hah2cdg59e6?MSPStoreType=image/gif&s=14&w=106&h=41[/URL] equals \sqrt{5}, deduce the values of a and b.

The Attempt at a Solution



So far, I've multiplied the numerator and denominator by the conjugate, and now I have [PLAIN]http://www4d.wolframalpha.com/Calculate/MSP/MSP231719ggb7fd1fig87b800001ie7f55ge9hc9190?MSPStoreType=image/gif&s=9&w=106&h=42. I've tried squaring the bottom and the top and expanding them out, but I'm just spinning my wheels at this point, I don't really know how to proceed. I would appreciate any hints you could give me, thanks in advance.

Are the fractions in the images correct? Because if I multiply
(\sqrt{a+bx}-\sqrt{5})(\sqrt{a+bx}+\sqrt{5})
I get
a + bx - 5,
and that's not what's in the numerator of the 2nd image.
 
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Sorry, I left out this part -[PLAIN]http://www4d.wolframalpha.com/Calculate/MSP/MSP157319ggbcc94ih84dd300005ia505e2i20325d5?MSPStoreType=image/gif&s=24&w=126&h=46 - that's what I got after multiplying by the conjugate, then I tried to cancel the x to get rid of the division by zero at the limit
 
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cripple said:
[PLAIN]http://www4d.wolframalpha.com/Calculate/MSP/MSP231719ggb7fd1fig87b800001ie7f55ge9hc9190?MSPStoreType=image/gif&s=9&w=106&h=42[/QUOTE]

Are you from University of Auckland? My friend from Auckland had the exact same problem. Anyways...

You multiplied incorrectly. After multiplying by the conjugate, you get:

(a+bx-5)/(x√(a+bx)+√5))

The only way this will approach a limit is if you manage to cancel out the x on the bottom. The only way to do that is to get the fraction to have the form x/x.
 
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You mean that you tried to cancel the x going from this:
[PLAIN]http://www4d.wolframalpha.com/Calculate/MSP/MSP157319ggbcc94ih84dd300005ia505e2i20325d5?MSPStoreType=image/gif&s=24&w=126&h=46
to this?
[PLAIN]http://www4d.wolframalpha.com/Calculate/MSP/MSP231719ggb7fd1fig87b800001ie7f55ge9hc9190?MSPStoreType=image/gif&s=9&w=106&h=42
If so, then you can't do that.
 
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Harrisonized said:
Are you from University of Auckland? My friend from Auckland had the exact same problem. Anyways...

You multiplied incorrectly. After multiplying by the conjugate, you get:

(a+bx-5)/(x√(a+bx)+√5))

The only way this will approach a limit is if you manage to cancel out the x on the bottom. The only way to do that is to get the fraction to have the form x/x.
I did this step, I should have included it in the OP. anyway yes I am from Auckland University, I am terrible at maths but decided to do 108 anyway, which is starting to look like it may have been a mistake:smile:
 


Thanks eumyang, I'll try again.
 


Hint (copied directly from your assignment): Multiply the numerator and denominator by the conjugate and remember that the denominator can not be zero.

The bolded part means that a=5.
 


Thanks Harrisonized, didn't realize the hint about the denominator meant a=5
 
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So how do we deduce that a=5 from that? I am at this step: (a+bx-5)/(x√(a+bx)+√5))
but unsure how to move forward from there.
 
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