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

2 Answers
George C.
Nov 9, 2016

#lim_(x->5^-) (x-5)/(x^2-25) = 1/10#

Explanation:

#lim_(x->5^-) (x-5)/(x^2-25) = lim_(x->5^-) color(red)(cancel(color(black)((x-5))))/(color(red)(cancel(color(black)((x-5))))(x+5))#

#color(white)(lim_(x->5^-) (x-5)/(x^2-25)) = lim_(x->5^-) 1/(x+5)#

#color(white)(lim_(x->5^-) (x-5)/(x^2-25)) = 1/(5+5)#

#color(white)(lim_(x->5^-) (x-5)/(x^2-25)) = 1/10#

Mia
Nov 9, 2016

#1/10#

Explanation:

Substituting #5 # in the given expression results to an indeterminate solution:

#lim_(x->5^-)(x-5)/(x^2-25)#

#=(5-5)/(5^2-25)#

#=0/0" "# Indeterminate solution.

let us think about another way to find the limit.

Here, Factorizing #x^2 - 25# is the best way.

Factorization #x^2 -25" "#is computed by applying the

polynomial identity.

#color(blue)((a^2-b^2)=(a-b)(a+b))#

#color(blue)(x^2 - 25=x^2 - 5^2=(x-5)(x+5))#

#lim_(x->5^-)(x-5)/(x^2-25)#

#=lim_(x->5^-)cancel(x-5)/color(blue)(cancel(x-5)(x+5))#

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

#=1/(5 + 5)#

#=1/10#

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