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

Question

Isabella Ackerman · Accepted Answer

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

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#

Charles Arocha · Answer

undefined To find the limit of (2 - √(3x)) / (√(x - 4) - 2) as x approaches 8, we can use algebraic manipulation and the concept of conjugate pairs.

First, we notice that both the numerator and denominator involve square roots. To eliminate the square roots, we can multiply both the numerator and denominator by the conjugate of the denominator, which is (√(x - 4) + 2).

By multiplying the numerator and denominator by (√(x - 4) + 2), we get:

[(2 - √(3x)) * (√(x - 4) + 2)] / [(√(x - 4) - 2) * (√(x - 4) + 2)]

Expanding and simplifying this expression, we have:

[(2√(x - 4) - √(3x)(√(x - 4)) + 4 - 2√(3x))] / [(x - 4) - 4]

Further simplifying, we get:

[(2√(x - 4) - √(3x)(√(x - 4)) + 4 - 2√(3x))] / (x - 8)

Now, we can cancel out the common terms in the numerator:

[2√(x - 4) - √(3x)(√(x - 4)) - 2√(3x)] / (x - 8)

As x approaches 8, we substitute this value into the expression:

[2√(8 - 4) - √(3 * 8)(√(8 - 4)) - 2√(3 * 8)] / (8 - 8)

Simplifying further:

[2√4 - √24 - 2√24] / 0

Since the denominator is 0, we cannot directly evaluate the limit using this method. We need to try a different approach, such as L'Hôpital's Rule or factoring and simplifying the expression further.

Christopher Allers · Answer

undefined To find the limit of the given expression as $ x $ approaches 8, you can rationalize the expression by multiplying both the numerator and the denominator by the conjugate of the denominator. This will help eliminate the radical in the denominator. After rationalizing, you can then substitute $ x = 8 $ into the expression and evaluate the limit.

$$ \lim_{x \to 8} \frac{2 - \sqrt{3x}}{\sqrt{x - 4} - 2} $$

Rationalizing the expression:

$$ \lim_{x \to 8} \frac{(2 - \sqrt{3x})(\sqrt{x - 4} + 2)}{(\sqrt{x - 4} - 2)(\sqrt{x - 4} + 2)} $$

$$ = \lim_{x \to 8} \frac{(2 - \sqrt{3x})(\sqrt{x - 4} + 2)}{x - 4 - 4} $$

$$ = \lim_{x \to 8} \frac{(2 - \sqrt{3x})(\sqrt{x - 4} + 2)}{x - 8} $$

$$ = \frac{(2 - \sqrt{3 \cdot 8})(\sqrt{8 - 4} + 2)}{8 - 8} $$

$$ = \frac{(2 - \sqrt{24})(\sqrt{4} + 2)}{0} $$

$$ = \frac{(2 - 2\sqrt{6})(2 + 2)}{0} $$

$$ = \frac{4 - 4\sqrt{6}}{0} $$

As $ x $ approaches 8, the denominator approaches 0. Hence, the limit is undefined or diverges.