Question #becac

1 Answer
sente
Oct 24, 2016

#lim_(x->-1)(sqrt(x^2+8)-3)/(x+1)=-1/3#

Explanation:

As we are getting a #0# in the numerator and the denominator, on direct substitution, our strategy will be to rationalize the numerator by using the identity #(a+b)(a-b) = a^2-b^2#. When we have two polynomial expressions with #-1# as a root, we will be able to cancel the #x+1# factors.

#lim_(x->-1)(sqrt(x^2+8)-3)/(x+1) = lim_(x->-1)((sqrt(x^2+8)-3)(sqrt(x^2+8)+3))/((x+1)(sqrt(x^2+8)+3))#

#=lim_(x->-1)(x^2+8-9)/((x+1)(sqrt(x^2+8)+3))#

#=lim_(x->-1)(x^2-1)/((x+1)(sqrt(x^2+8)+3))#

#=lim_(x->-1)((x+1)(x-1))/((x+1)(sqrt(x^2+8)+3))#

#=lim_(x->-1)(x-1)/(sqrt(x^2+8)+3)#

#=(-1-1)/(sqrt((-1)^2+8)+3)#

#=-1/3#

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