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How do you find the limit of #(abs(x^2-25))/(x+5)# as x approaches -5?

Answer 1

Isabella Ackerman

This limit does not exists.

#(abs(x^2-25))/(x+5) = (abs(x+5)abs(x-5))/(x+5) = "signum"(x+5)abs(x-5)#

We have two options

#lim_{x->-5^-}"signum"(x+5)abs(x-5) = -10# and #lim_{x->-5^+}"signum"(x+5)abs(x-5) = 10# so this limit does not exists

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Answer 2

Isaiah Allen

To find the limit of (abs(x^2-25))/(x+5) as x approaches -5, we can evaluate the expression by substituting -5 for x. This gives us (abs((-5)^2-25))/((-5)+5), which simplifies to (abs(0))/0. Since the absolute value of 0 is 0, the expression becomes 0/0. This is an indeterminate form, so we need to further simplify the expression. By factoring the numerator, we get abs((x-5)(x+5))/(x+5). Now, we can cancel out the common factor of (x+5) in the numerator and denominator, resulting in abs(x-5). Finally, as x approaches -5, the absolute value of (x-5) approaches 10. Therefore, the limit of (abs(x^2-25))/(x+5) as x approaches -5 is 10.

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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.