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

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The problem involves finding constants a and b such that the limit as x approaches 1 of the expression (a√(x+3) - b) / (x - 1) equals 1. To solve this, one must recognize that both the numerator and denominator approach zero as x approaches 1, necessitating the application of L'Hôpital's Rule. By differentiating the numerator and denominator, and substituting x = 1, a linear relationship between a and b can be established, allowing for the determination of both constants.

PREREQUISITES
  • Understanding of limits in calculus
  • Familiarity with L'Hôpital's Rule
  • Knowledge of differentiation techniques
  • Ability to manipulate algebraic expressions involving square roots
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  • Study the application of L'Hôpital's Rule in depth
  • Practice solving limit problems involving indeterminate forms
  • Explore differentiation of composite functions
  • Review algebraic techniques for simplifying expressions with square roots
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Students and educators in calculus, mathematicians dealing with limits, and anyone interested in solving algebraic limit problems involving constants.

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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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Is this your problem?

[tex]\lim_{x\rightarrow{1}} \frac{a\sqrt{x+3}-b}{x-1}=1[/tex]?
 
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.
 

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