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Calculus/Trig Limit solving with a square root

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Question
Hi- I am taking AP Calc BC, and I understand most of the work, but I am having trouble with solving for a limit using only the Limit Laws when there is a square root involved. The tricky question for me is:
        x
lim   --------
x->0  √(1+3x)-1

Thank you!

Answer
If we multiply the top and the bottom of the fraction by √(1+3x) + 1,
we get (x(√(1+3x) + 1)/[(√(1+3x) + 1)(√(1+3x) - 1)].

Note that the bottom is (a+b)*(a-b), which is aČ - bČ.
See that aČ=1+3x, bČ=1, so aČ-bČ= 3x.

The top is x√(1+3x) + 1).

The top and bottom can both be divided by x (since x=0 is only a limit - we're not actually there), and the answer can be obtained by putting in x = 0.

Thanks for the question and have a great day.

Calculus

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