Question

How do you find the limit of `(2 - root3x) / (sqrt(x - 4) - 2)` as x approaches 8?

Answer 1

Use the conjugates of the numerator and the denominator to find that the limit is `-1/3`.

Explanation:

The conjugate of `sqrtu -v` is `sqrtu + v` because the product, `u-v^2` has no radicals. (That is not the definition of conjugate, but it works for this question.)

We can understand why this works by recalling from more basic algebra, that `(a-b)(a+b) = a^2-b^2`. So if one, or both, of `a`, `b` involve a square (second) root, then their squares do not involve roots.

Consider, then, third powers and third roots.

I hope that you learned in your previous study of algebra that

`a^3-b^3= (a-b)(a^2+ab+b^2)` and `a^3+b^3= (a+b)(a^2-ab+b^2)`.

Here, if `a` or `b` or both involve cube (third) roots, then the cubes do not involve roots.

On to the question at hand:

`lim_(xrarr8)(2 - root3x) / (sqrt(x - 4) - 2)`

If we try to evaluate by substitution, we get the indeterminate form `0/0`.
We'll use the conjugates to rewrite the quotient:

`(2 - root3x) / (sqrt(x - 4) - 2) * ((2^2+2root3x+root3(x^2)))/((2^2+2root3x+root3(x^2)))* ((sqrt(x - 4) + 2))/((sqrt(x - 4) + 2))`

`= ((2^3-x)(sqrt(x - 4) +2))/(((x-4)-4)(2^2+2root3x+root3(x^2)))`

`= ((8-x)(sqrt(x - 4) +2))/((x-8)(2^2+2root3x+root3(x^2)))`

`= (-1(sqrt(x - 4) +2))/((2^2+2root3x+root3(x^2)))`.

We can now evaluate the limit by substitution. The form is no longer indeterminate.

`lim_(xrarr8)(2 - root3x) / (sqrt(x - 4) - 2) = lim_(xrarr8)(-1(sqrt(x - 4) +2))/((2^2+2root3x+root3(x^2)))`

`= (-1(sqrt4+2))/(4+4+4) = -1/3`