MHB Solving Limit as x approaches 3 using Multiplication and Division

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To solve the limit as x approaches 3 for the expression involving square roots, it's suggested to multiply both the numerator and denominator by the conjugate of the numerator, which is $\sqrt{6x - 14} + \sqrt{x + 1}$. This approach helps eliminate the square roots and simplifies the expression. The discussion emphasizes that x - 3 is indeed a factor in the numerator, and making it explicit aids in the simplification process. Ultimately, this method leads to a clearer path for evaluating the limit. The conversation concludes with the participant expressing understanding of the solution.
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I have to solve this limit.

$$\lim_{{x}\to{3}} \frac{\sqrt{6x - 14} - \sqrt{x + 1}}{x -3}$$

Now, I think that by definition x - 3 is a divisor of the numerator, but how do I advance from here? Do I do long division?
 
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tmt said:
I have to solve this limit.

$$\lim_{{x}\to{3}} \frac{\sqrt{6x - 14} - \sqrt{x + 1}}{x -3}$$

Now, I think that by definition x - 3 is a divisor of the numerator, but how do I advance from here? Do I do long division?

How about making the x - 3 in the numerator explicit?
What do you get if you multiply both numerator and denominator by $\sqrt{6x - 14} + \sqrt{x + 1}$?
 
I like Serena said:
How about making the x - 3 in the numerator explicit?
What do you get if you multiply both numerator and denominator by $\sqrt{6x - 14} + \sqrt{x + 1}$?

Okay thanks, now I got it.
 

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