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

Answer 1

Cameron Alexander

#1/10#

#(x-5)/(x^2-5)=(x-5)/((x-5)(x+5))=1/(x+5), " for each " x!=5#

#lim_(x->5) f(x) = lim_(x->5) 1/(x+5) = 1/10#

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Answer 2

Christopher Agresta

To find the limit of (x-5)/(x^2-25) as x approaches 5, we can simplify the expression by factoring the denominator. The denominator can be factored as (x-5)(x+5). Canceling out the common factor of (x-5), we are left with 1/(x+5).

Now, substituting x=5 into the simplified expression, we get 1/(5+5) = 1/10.

Therefore, the limit of (x-5)/(x^2-25) as x approaches 5 is 1/10.

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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.

How do you find the limit of #(x-5)/(x^2-25)# as #x-&gt;5#?

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