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

Question

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

Aurora Alvarado · Accepted Answer

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#

Ezekiel Alexander · Answer

undefined To find the limit of [root3(x) - 5] / (x-125) as x approaches 125, we can use algebraic manipulation and the concept of limits.

First, let's simplify the expression by rationalizing the numerator. We can multiply both the numerator and denominator by the conjugate of the numerator, which is [root3(x) + 5].

This gives us [(root3(x) - 5) * (root3(x) + 5)] / [(x-125) * (root3(x) + 5)].

Expanding the numerator using the difference of squares formula, we get [root3(x)^2 - 5^2] / [(x-125) * (root3(x) + 5)].

Simplifying further, we have [x - 25] / [(x-125) * (root3(x) + 5)].

Now, we can evaluate the limit as x approaches 125.

Substituting x = 125 into the expression, we get [125 - 25] / [(125-125) * (root3(125) + 5)].

Simplifying this, we have 100 / (0 * (root3(125) + 5)).

Since the denominator is 0, the limit does not exist.