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Why eliminate answers?

  • Thread starter razored
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  • #1
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[SOLVED] Why eliminate answers?

In my book, Essential Calculus, a section is introduced with limits. They introduce a random(maybe) function :

[tex]Q(h) = \frac {2h + h^2}{h} (h \neq 0)[/tex]

"We then divide the numerator by the denominator h, which is permissible since [tex]h \neq 0[/tex]. This gives the simple formula [tex]Q(h) = 2 + h (h \neq 0)[/tex]"

I was always told you do not divide by variables like in a trigonometric equation because it eliminates solutions. How are they then to say that it is permissible since [tex]h \neq 0[/tex] ? I don't understand.
 

Answers and Replies

  • #2
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In this case, it is fine. Given that [tex]h \neq 0[/tex], both functions are equal because you are only simplifying the function.
 
  • #3
tiny-tim
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I was always told you do not divide by variables like in a trigonometric equation because it eliminates solutions. How are they then to say that it is permissible since [tex]h \neq 0[/tex] ? I don't understand.
Hi razored! :smile:

It's ok because all these calculus equations begin "lim as h -> 0".

So it's impossible for h to be 0. :smile:

(oh … and look up L'Hôpital's Rule :wink:)
 
  • #4
HallsofIvy
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In my book, Essential Calculus, a section is introduced with limits. They introduce a random(maybe) function :

[tex]Q(h) = \frac {2h + h^2}{h} (h \neq 0)[/tex]

"We then divide the numerator by the denominator h, which is permissible since [tex]h \neq 0[/tex]. This gives the simple formula [tex]Q(h) = 2 + h (h \neq 0)[/tex]"

I was always told you do not divide by variables like in a trigonometric equation because it eliminates solutions. How are they then to say that it is permissible since [tex]h \neq 0[/tex] ? I don't understand.
What "solutions" do you mean? You are not "solving" an equation here.
 
  • #5
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What "solutions" do you mean? You are not "solving" an equation here.
Whoops. That is what I misunderstood. Thanks for pointing that out!
 

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