Limits
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HallsofIvy
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For the first problem, \(\displaystyle \lim_{x\to 4} \frac{x- 4}{x^2- 3x- 4}\)
Your very first thought should be to replace "x" with "4".
Unfortunately, the denominator becomes \(\displaystyle 4^2- 3(4)- 4=16-12- 4= 0\) so we cannot do that. IF the numerator were not also 0 we could simply annouce there is no limit.
However the numerator becomes 4- 4= 0. But the very fact that x= 4 makes \(\displaystyle x^2- 3x- 4= 0\) means that x- 4 is a factor of \(\displaystyle x^3- 3x- 4\). Once we realize that \(\displaystyle x^2- 3x- 4= (x- 4)(x- a)\) it is easy to see that "a" must be -1 so that \(\displaystyle \frac{x- 4}{x^2- 3x- 4}= \frac{x- 4}{(x- 4)(x+1)}\).
As long as x is not 4, so that x- 4 is not 0, we can cancel- \(\displaystyle \frac{x- 4}{x^2- 3x- 4}= \frac{x- 4}{(x- 4)(x+1)}= \frac{1}{x+1}\). Strictly speaking \(\displaystyle \frac{x- 4}{x^2- 3x- 4}\) is NOT the same as \(\displaystyle \frac{1}{x+1}\) (one is not defined at x= 4 while the other is 1/5) but since the "limit as x goes to 4" does not depend on a value at x= 4, the limits are the same- \(\displaystyle \lim_{x\to 4}\frac{x- 4}{x^2- 3x- 4}= \lim_{x\to 4}\frac{1}{x+ 1}= \frac{1}{5}\).
Your very first thought should be to replace "x" with "4".
Unfortunately, the denominator becomes \(\displaystyle 4^2- 3(4)- 4=16-12- 4= 0\) so we cannot do that. IF the numerator were not also 0 we could simply annouce there is no limit.
However the numerator becomes 4- 4= 0. But the very fact that x= 4 makes \(\displaystyle x^2- 3x- 4= 0\) means that x- 4 is a factor of \(\displaystyle x^3- 3x- 4\). Once we realize that \(\displaystyle x^2- 3x- 4= (x- 4)(x- a)\) it is easy to see that "a" must be -1 so that \(\displaystyle \frac{x- 4}{x^2- 3x- 4}= \frac{x- 4}{(x- 4)(x+1)}\).
As long as x is not 4, so that x- 4 is not 0, we can cancel- \(\displaystyle \frac{x- 4}{x^2- 3x- 4}= \frac{x- 4}{(x- 4)(x+1)}= \frac{1}{x+1}\). Strictly speaking \(\displaystyle \frac{x- 4}{x^2- 3x- 4}\) is NOT the same as \(\displaystyle \frac{1}{x+1}\) (one is not defined at x= 4 while the other is 1/5) but since the "limit as x goes to 4" does not depend on a value at x= 4, the limits are the same- \(\displaystyle \lim_{x\to 4}\frac{x- 4}{x^2- 3x- 4}= \lim_{x\to 4}\frac{1}{x+ 1}= \frac{1}{5}\).