How do you find the limit $lim_(x->3^+)|3-x|/(x^2-2x-3)$ ?

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Answer 1 · 2014-10-05T13:19:07

$lim_{x to 3^+}{|3-x|}/{x^2-2x-3}$

Note that since $x$ approaches $3$ from the right, $(3-x)$ is negative, which means that $|3-x|=-(3-x)=x-3$ .

by removing the absolute value sign and factoring out the denominator,

$=lim_{x to 3^+}{x-3}/{(x-3)(x+1)}$

by cancelling out $(x-3)$ 's,

$=lim_{x to 3^+}1/{x+1}=1/{3+1}=1/4$

I hope that this was helpful.

How do you find the limit lim_(x->3^+)|3-x|/(x^2-2x-3)lim_(x->3^+)|3-x|/(x^2-2x-3) ?