How do you find the limit of #(x²-25)/(sqrt(2x+6)-4)# as x approaches 5?

1 Answer
Truong-Son N.
Jul 30, 2016

In this case, we really have a #0/0# form, an indeterminate form, so we can use L'Hopital's rule, where we take the derivative of the numerator and denominator separately and retain the same limit.

#color(blue)(lim_(x->5) (x^2 - 25)/(sqrt(2x + 6) - 4))#

#= lim_(x->5) [d/(dx)(x^2 - 25)]/[d/(dx)(sqrt(2x + 6) - 4)]#

#= lim_(x->5) [2x]/[1/(cancel(2)sqrt(2x + 6))*cancel(2)]#

#= lim_(x->5) 2xsqrt(2x + 6)#

#= 2(5)sqrt(2(5) + 6)#

#= 10*4 = color(blue)(40)#

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