How to solve this limit? (difference of two square rooted terms)

In summary, to solve a limit involving the difference of two square rooted terms, one can use the conjugate rule by multiplying the numerator and denominator of the fraction by the conjugate of the denominator, which is the same expression but with the opposite sign between the terms. This allows for the elimination of the square root in the denominator and simplification of the expression. However, the conjugate rule is not the only method to solve these types of limits, as other algebraic techniques can also be used. An example of solving a limit using the conjugate rule is provided, where the limit is evaluated to be -1/8 when x approaches 4.
  • #1
mnb96
715
5
Hello,
do you have a hint on how to solve this limit?

[tex]\lim_{x\to +\infty}(\sqrt{x^2+ax+b}-\sqrt{x^2-ax+b})[/tex]

Thanks!
 
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  • #2
When I see an expression of the form

[tex]\sqrt{A} - \sqrt{B}[/tex]

I think of multiplying it by

[tex]\frac{\sqrt{A} + \sqrt{B}}{\sqrt{A} + \sqrt{B}}[/tex]

Does that help?

Petek
 
  • #3
Thanks a lot!
that trick made the job. The limit of that expression is a
 

FAQ: How to solve this limit? (difference of two square rooted terms)

How do I solve this type of limit?

To solve a limit involving the difference of two square rooted terms, you can use the conjugate rule. This means multiplying the numerator and denominator of the fraction by the conjugate of the denominator, which is the same expression but with the opposite sign between the terms.

What is the conjugate of a square rooted term?

The conjugate of a square rooted term is the same expression but with the opposite sign between the terms. For example, the conjugate of √x + 2 is √x - 2.

Why do we use the conjugate rule to solve this type of limit?

The conjugate rule allows us to get rid of the square root in the denominator, making it easier to evaluate the limit. It also helps to simplify the expression and often leads to a solution.

Do I always have to use the conjugate rule to solve this type of limit?

No, the conjugate rule is not the only method to solve limits involving the difference of two square rooted terms. You can also try simplifying the expression by factoring or using other algebraic techniques.

Can you provide an example of solving a limit using the conjugate rule?

Sure, let's say we have the limit √x - √4 / x - 4 as x approaches 4. Using the conjugate rule, we multiply the numerator and denominator by the conjugate of the denominator, which is √x + 2. This gives us (√x - √4)(√x + 2) / (x - 4)(√x + 2). We can then simplify this expression to (√x^2 - 2√x - 4) / (x^2 - 16). Now, we can substitute x = 4 into this simplified expression and find that the limit is equal to -1/8.

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