Home > Calculus > Determining Limits Algebraically

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

Answer 1

Emily Ammerman

For #x>3#, we can write #|3-x|/{x^2-2x-3}={x-3}/{(x-3)(x+1)}=1/{x+1}#

So, #lim_{x to 3^+}|3-x|/{x^2-2x-3} =lim_{x to 3^+}1/{x+1}=1/{3+1}=1/4#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Evelyn Agard

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.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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

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