Solving a Math Problem with Constants and Limits: Find a and b

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Given a, b are constants and lim x approch 1, a root x+3 - b all over x-1 equals 1. Find a and b

No clue how to answer this, but this is what I think i can do

a) get rid of the sqrt
b) apply limit

See, the problem is I have to unknows so i don't know what to do
 
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\lim_{x\rightarrow{1}} \frac{a\sqrt{x+3}-b}{x-1}=1?
 
gator said:
Given a, b are constants and lim x approch 1, a root x+3 - b all over x-1 equals 1. Find a and b

No clue how to answer this, but this is what I think i can do

a) get rid of the sqrt
b) apply limit

See, the problem is I have to unknows so i don't know what to do

If the version posted in LaTex is correct, note that the denominator goes to zero when x->1. For the quotient to have a finite limit, the numerator must also go to zero when x is set to 1.

You get a simple linear relationship between a and b. ---(1)

Now you can use L'Hopital's Rule to evaluate a limit of the form 0/0. So differentiate both numerator and denominator. You know that the quotient of these two is also going to be 1 at the limit x->1.

So set x = 1 in that. The b term would have vanished, so you can now solve for a. Put that back in equation (1) and work out b, you're done.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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