Why eliminate answers?

[razored]

[SOLVED] Why eliminate answers?

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

$$Q(h) = \frac {2h + h^2}{h} (h \neq 0)$$

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

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 $$h \neq 0$$ ? I don't understand.

 

[konthelion]

In this case, it is fine. Given that $$h \neq 0$$, both functions are equal because you are only simplifying the function.

 

[tiny-tim]

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 $$h \neq 0$$ ? I don't understand.

Hi razored!

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

So it's impossible for h to be 0.

(oh … and look up L'Hôpital's Rule )

 

[HallsofIvy]

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

$$Q(h) = \frac {2h + h^2}{h} (h \neq 0)$$

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

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 $$h \neq 0$$ ? I don't understand.

What "solutions" do you mean? You are not "solving" an equation here.

 

[razored]

What "solutions" do you mean? You are not "solving" an equation here.

Whoops. That is what I misunderstood. Thanks for pointing that out!