Method

Dividing Out Technique for Limits

The dividing out technique is used to evaluate limits when the direct substitution leads to the indeterminate form

\frac{0}{0} .

This method involves identifying and canceling common factors between the numerator and the denominator of a rational function. The limit value can then be directly substituted into the simplified form of the function to evaluate it. For example, consider the following function.

x \to 1 lim \frac{2 x ^{2} - 8 x + 6}{x - 1}

x

approaches

1,

both the numerator and denominator of this function become

0,

so the dividing out technique can be applied to find the limit of the function. This method involves three main steps.

Factor the Expressions

Factor both the numerator and the denominator of the expression, if possible. For the given function, only the numerator can be factored.

2 x^{2} - 8 x + 6 = 2 (x - 3) (x - 1)

The function can now be rewritten using the factored form of the numerator.

x \to 1 lim \frac{2 x ^{2} - 8 x + 6}{x - 1} = x \to 1 lim \frac{2 ( x - 3 ) ( x - 1 )}{x - 1}

Divide Out Common Factors

Divide out common factors that appear in both the numerator and the denominator. In other words, cancel out common factors. The expression found in Step

1

shows that the only common factor between the numerator and the denominator of the example function is

(x - 1) .

x \to 1 lim \frac{2 ( x - 3 ) ( x - 1 )}{x - 1}

CancelCommonFac

Cancel out common factors

x \to 1 lim \frac{2 ( x - 3 ) ( x - 1 )}{( x - 1 )}

SimpQuot

Simplify quotient

x \to 1 lim 2 (x - 3)

It is always important to simplify the resulting expression whenever possible.

Apply Direct Substitution

Direct substitution can now be used to determine the limit of the simplified function. It is important to note that the limit of the simplified function must be equal to the limit of the original function. For the given example,

1

must be substituted for

x

into the simplified form to find the limit of the original function.

x \to 1 lim \frac{2 x ^{2} - 8 x + 6}{x - 1} = x \to 1 lim 2 (x - 3)

Substitute

$x = 1$

x \to 1 lim \frac{2 x ^{2} - 8 x + 6}{x - 1} = 2 (1 - 3)

SubTerm

Subtract term

x \to 1 lim \frac{2 x ^{2} - 8 x + 6}{x - 1} = 2 (- 2)

MultPosNeg

$a (- b) = - a \cdot b$

x \to 1 lim \frac{2 x ^{2} - 8 x + 6}{x - 1} = - 4

Note that if the expression still shows an indeterminate form, it is possible that there are more common factors that were not divided out.

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