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

By factoring out the denominator and eliminating the absolute value sign,

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To find the limit of lim_(x->3^+)|3-x|/(x^2-2x-3), we can evaluate the limit from the right side of x=3.

First, let's simplify the expression. The absolute value of (3-x) is equal to (x-3) when x>3. So, we can rewrite the expression as (x-3)/(x^2-2x-3).

Next, we factor the denominator. The expression becomes (x-3)/[(x-3)(x+1)].

Now, we can cancel out the common factor of (x-3) in the numerator and denominator. This leaves us with 1/(x+1).

Finally, we can take the limit as x approaches 3 from the right side. Plugging in x=3 into the simplified expression, we get 1/(3+1) = 1/4.

Therefore, the limit of lim_(x->3^+)|3-x|/(x^2-2x-3) is 1/4.

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To find the limit (\lim_{x \to 3^+} \frac{|3 - x|}{x^2 - 2x - 3}), first consider values of (x) approaching 3 from the right side ((x \to 3^+)).

As (x) approaches 3 from the right side, the expression (|3 - x|) becomes (|3 - 3|) which equals 0.

Also, as (x) approaches 3 from the right side, the expression (x^2 - 2x - 3) approaches (3^2 - 2(3) - 3 = 0).

Therefore, when evaluating the limit, you have (0/0), which is an indeterminate form.

To evaluate further, factorize the denominator (x^2 - 2x - 3 = (x - 3)(x + 1)).

Now, you can simplify the expression:

[ \lim_{x \to 3^+} \frac{|3 - x|}{x^2 - 2x - 3} = \lim_{x \to 3^+} \frac{0}{(x - 3)(x + 1)} = \lim_{x \to 3^+} \frac{0}{0} ]

Since you still have an indeterminate form, you can apply L'Hôpital's Rule by differentiating the numerator and the denominator with respect to (x).

After applying L'Hôpital's Rule, you get:

[ \lim_{x \to 3^+} \frac{|3 - x|}{x^2 - 2x - 3} = \lim_{x \to 3^+} \frac{-1}{2x - 2} ]

Now, plug in (x = 3) into the expression to find the limit:

[ \lim_{x \to 3^+} \frac{-1}{2x - 2} = \frac{-1}{2(3) - 2} = \frac{-1}{4} ]

Therefore, the limit (\lim_{x \to 3^+} \frac{|3 - x|}{x^2 - 2x - 3}) is (-\frac{1}{4}).

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