I'm reviewing limits to tutor a student in precalc and came across a problem.(adsbygoogle = window.adsbygoogle || []).push({});

The conjugate method multiplies the numerator and denominator by the conjugate of the numerator or denominator to simplify the equation. However, after a quick example I wrote for myself, I found that:

lim x-> 3

f(x) / (x-3)

Conjugate Method on the Denominator

lim x-> 3

f(x)*(x+3) / (x^2-9)

The limit still has a 0 in the denominator

lim x-> -3

f(x) / (x+3)

Conjugate Method on the Denominator

lim x-> -3

f(x)*(x-3) / (x^2-9)

The limit still has a 0 in the denominator

The examples of "The Conjugate Method" that I've found online have all multiplied by the conjugate to create a common factor to eliminate the 0 in the denominator.

Ex.

lim x-> 4

(sqrt(x)-2) / x-4

Conjugate of Numerator

lim x-> 4

x-4 / ((x-4)(sqrt(x)+2)

Cancel common factor

lim x-> 4

1 / (sqrt(x)+2)

=1/4

My issue is that the conjugate method is the same as the factoring method as far as I understand it. Is there a difference or is the conjugate method simply easier to use when the factors aren't as easily identifiable.

My point using the example above is:

lim x-> 4

(sqrt(x)-2) / x-4

Factor the denominator into (sqrt(x)-2) & (sqrt(x)+2)

lim x-> 4

(sqrt(x)-2) / ((sqrt(x)-2)(sqrt(x)+2))

Cancel (sqrt(x)-2)

lim x-> 4

1 / (sqrt(x)+2)

= 1/4

Is there a difference between these methods?

Is there a time when only one or the other can be used?

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# The Conjugate Method for Limits

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