How do you evaluate the limit #((2x+8)/(x^2-12)-1/x)/(x+6)# as x approaches -6?

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Answer 1 · 2016-11-19T16:14:12

The limit is #1/36#.

#=>lim_(x-> -6)((x(2x + 8) - (1(x^2 - 12)))/(x(x^2 - 12)))/(x + 6)#

#=>lim_(x->-6)((2x^2 + 8x - x^2 + 12)/(x^3 - 12x))/(x + 6)#

#=>lim_(x->-6) ((x^2 + 8x + 12)/(x^3 - 12x))/(x + 6)#

#=>lim_(x->-6) (((x + 6)(x + 2))/(x^3 - 12x))/(x + 6)#

#=>lim_(x->-6) ((x + 6)(x+ 2))/((x^3 - 12x)(x + 6))#

#=>lim_(x->-6) (x + 2)/(x^3- 12x)#

#=> (-6 + 2)/(-6^3 - 12(-6))#

#=> (-4)/(-144)#

#=> 1/36#

Hopefully this helps!