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December 20, 2014

December 20, 2014

Posted by **Amy** on Wednesday, February 7, 2007 at 6:32pm.

The problem is:

Find the limit:

lim (x->5)

The numerator is (the square root of 9-x) -2 and the denominator is x-(the square root of x+20).

I know that I can't substitute 5 for x because it'll create 0/0, but other than making a table of values close to 5, I don't know what to do. My teacher says to rationalize the numerator and the denominator at the same time, but I have no clue as to what he's talking about.

Help?

Amy :)

You can manipulate the square roots, but this is not really one calculates such limits in practice. The practical way is to insert for x: 5 + epsilon and then expand in powers of epsilon.

In the case of a fraction, like this case, this reduces to L'Hopital's rule: Differentiate the numerator and denominator separately and then insert the limiting value.

If f(a) = g(a) = 0 then:

Lim x-->a f(x)/g(x) =

Lim h-->0 f(a+h)/g(a+h) =

Lim h-->0 [f(a+h) - f(a)]/[g(a+h)-g(a)]

Divide numerator and denominator by h inside the limit. The limit of a fraction is the ratio of the limits of the denominator and numerators separately, if that ratio exists.

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