How do you evaluate the limit of #lim (2x-8)/(x^3-64)# as #x->4#?

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Answer 1 · 2017-01-18T17:10:01

#Lt_(x->4)(2x-8)/(x^3-64)=1/24#

#Lt_(x->4)(2x-8)/(x^3-64)#

and as #x^3-64=x^3-4^3#,

using identity #(a^3-b^3)=(a-b)(a^2+ab+b^2)#, the above can be written as

= #Lt_(x->4)(2(x-4))/((x-4)(x^2+4x+16))#

= #Lt_(x->4)2/(x^2+4x+16)#

= #2/(4^2+4xx4+16)#

= #2/(16+16+16)#

= #1/24#