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charley076
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I'm reviewing limits to tutor a student in precalc and came across a problem.
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?
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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