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How do you simplify and state the excluded values for #(x+5) /(x^2-25)#?

Answer 1

Note that #(x+5)/(x^2-25)# is only defined if the denominator is not equal to zero. That is #x^2-25 != 0# #rarr x=5# and #x=-5# are excluded values.

If #x!=+-5# then #(x+5)/(x^2-25)#

#= (x+5)/((x+5)(x-5))#

#= cancel(x+5)/(cancel(x+5)(x-5))#

#= 1/(x-5)#

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Answer 2

Ethan Adkins

To simplify the expression (x+5) /(x^2-25), we can factor the denominator as (x+5)(x-5). This allows us to cancel out the common factor of (x+5) in the numerator and denominator. The simplified expression is 1/(x-5).

To find the excluded values, we set the denominator equal to zero and solve for x. In this case, x^2-25 = 0. By factoring, we get (x+5)(x-5) = 0. Therefore, the excluded values are x = -5 and x = 5.

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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.