How do I determine the value of b so that each limit exists?

$\displaystyle \lim_{x \to b}

\frac{4x^2 + 11x -3}{x-b}

$

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- Mar 22nd 2010, 10:34 AMNeconineDetermine the value of b?
How do I determine the value of b so that each limit exists?

$\displaystyle \lim_{x \to b}

\frac{4x^2 + 11x -3}{x-b}

$ - Mar 22nd 2010, 12:14 PMdrumist
First you should factor the numerator:

$\displaystyle \lim_{x \to b} \frac{4x^2+11x-3}{x-b} = \lim_{x \to b} \frac{(4x-1)(x+3)}{x-b}$

Now, we should recognize that one of two things can happen from the term in the denominator. There will either be a removable discontinuity or an infinite discontinuity. If the denominator term will cancel with a term in the numerator, that means it will be a removable discontinuity. Otherwise it will be infinite.

Notice that at a removable discontinuity, the limit exists, but it doesn't exist at an infinite discontinuity.

So there are two values of $\displaystyle b$ that will cause there to be a removable discontinuity.