How do you find the limit of #[root3(x) - 5] / (x-125)# as x approaches 125?

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Answer 1 · 2016-08-13T17:00:07

Use #a^3-b^3 = (a-b)(a^2+ab+b^2)# to either rationalize the numerator or factor the denominator,

Method 1

#(root(3)(x)-5)/(x-125) * (root(3)(x)^2+5root(3)x+5^2)/(root(3)(x)^2+5root(3)x+5^2) = (x-125)/((x-125)(root(3)(x)^2+5root(3)x+5^2)) #

# = 1/(root(3)(x)^2+5root(3)x+5^2)#

Method 2

#(root(3)(x)-5)/(x-125) = (root(3)(x)-5)/((root(3)(x)-5)(root(3)(x)^2+5root(3)x+5^2)#

# = 1/(root(3)(x)^2+5root(3)x+5^2)#

Using either method we get to

#lim_(xrarr125) 1/(root(3)(x)^2+5root(3)x+5^2) = 1/(25+25+25)=1/75#