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Limits Involving Roots

April 8th, 2009 |

Author: Merrin

It is often helpful when calculating limits involving roots to multiply by a conjugate expression. What is a conjugate expression? In complex numbers, when we divide two complex numbers, we can use a conjugate expression of the form 1/1 to convert the expression to standard form.

$\left( \cfrac{a+ib}{c+id} \right) \left(\cfrac{c-id}{c-id} \right)=\cfrac{ac-bd}{c^2+d^2}+i\cfrac{bc-ad}{c^2+d^2}$

Conjugate expressions can be used in algebra when dealing with the difference of square roots in the denominator in order to rationalize the denominator.

Example 1. Rationalize the following function

$f(x) = \cfrac{1}{\sqrt{x+a} -\sqrt{x+b} }$

Solution 1. The conjugate expression for this function is a factor of the form 1/1

$\cfrac{1}{1}=\cfrac{\sqrt{x+a}+\sqrt{x+b}}{\sqrt{x+a}+\sqrt{x+b}}$

Multiplying this out, rationalizes the denominator

$\cfrac{1}{\sqrt{x+a} -\sqrt{x+b} } \left(\cfrac{\sqrt{x+a}+\sqrt{x+b}}{\sqrt{x+a}+\sqrt{x+b}}\right)= \cfrac{\sqrt{x+a}+\sqrt{x+b}}{a-b}$

Example 2. Calculate the following limit

$\mathop{\lim \limits_{ x\to \infty}} (\sqrt{x+a} -\sqrt{x+b} )\sqrt{x+c}$

Solution 2. The above limit is of an indeterminate form infinity minus infinity. Multiplying by the conjugate expression to the first factor simplifies the calculation.

$L = \mathop{\lim (}\limits_{x\to \infty } \sqrt{x+a} -\sqrt{x+b} )\sqrt{x+c} \\ L= \mathop{\lim (}\limits_{x\to \infty } \sqrt{x+a}-\sqrt{x+b})\left(\cfrac{\sqrt{x+a}+\sqrt{x+b}}{\sqrt{x+a}+\sqrt{x+b}} \right)\sqrt{x+c}\\ L=\mathop{\lim}\limits_{x\to \infty }(a-b) \cfrac{\sqrt{x+c}}{\sqrt{x+a}\sqrt{x+b}}= \cfrac{a-b}{2}$

Example 3. Find the following limit for the function

$f(x)=\sqrt{x}$

$\mathop{\lim }\limits_{h\to 0} \cfrac{f(x)-f(x-h)}{h}$

Solution 3. First we fill in the details of the function into the limit then we use a conjugate expression to rationalize the numerator which simplifies the calculation.

${L}= {\mathop{\lim }\limits_{h\to 0} \cfrac{f(x)-f(x-h)}{h} =\mathop{\lim }\limits_{h\to 0} \cfrac{\sqrt{x} -\sqrt{x-h} }{h} } \\ = {\mathop{\lim }\limits_{h\to 0} \cfrac{\sqrt{x} -\sqrt{x-h} }{h} \left(\cfrac{\sqrt{x} +\sqrt{x-h} }{\sqrt{x} +\sqrt{x-h} } \right)} \\ = {\mathop{\lim }\limits_{h\to 0} \cfrac{x-x+h}{h} \cfrac{1}{\sqrt{x} +\sqrt{x-h} } =\mathop{\lim }\limits_{h\to 0} \cfrac{1}{\sqrt{x} +\sqrt{x-h} } =\cfrac{1}{2\sqrt{x} } }$

Different expressions such as the difference of cube roots, fourth roots, and so on can also be rationalized, but for cube roots this requires figuring out three terms to multiply by instead of two. These terms can be found from an expression such as the factorization of the difference of cubes. Recall your knowledge of factoring to proceed.

${x^{3}-y^{3}=(x-y)(x^{2} +xy+y^{2} )} \\ {\cfrac{1}{x-y} =\cfrac{x^{2} +xy+y^{2} }{x^{3} -y^{3} } }$