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Calculus/limit of a square root

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Expert: - 10/12/2011


Question
I don't understand what to do with this question.
find the lim as x goes to 4 of (the square root of x plus 5) minus three, (that is the numerator) divided by x-4 (the denominator). I tried to multiply it by (the square root of x plus 5) plus three, over this same number so it would be like multiplying it by 1, but it came out to 1/3 which was not right. Please help me!

Answer
This sounds like find lim(x->4)[√(x+5) - 3)/(x-4).
When x=4 is put in, we get (√9 - 3)/(4-4) = 0/0.

Multiply the top and bottom of the fraction by (√(x+5) +3).
This gives (x+5 - 9)/[(x-4)(√(x+5) +3).

The numerator simplies to x-4 since 5-9=-4.
This gives an (x-4)/(x-4), which cancels.

What is left is 1/(√(x+5) +3).
Putting in x=4 gives 1/(√9 + 3) = 1/6.

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