Find the derivative of ?(?)=3 from first principles.

Finding the derivative of ?(?)=3 from first principles can be done as said in the posts above, but first just check that, in fact, the question says f(x)=3 and not something else.

Please respond and we can see how you are going with it.
 
I have a question for you. Since 0/0 is not 0, why is the limit that you wrote equals 0?
 
pka said:
Surely you know that in evaluating [imath]\mathop {\lim }\limits_{x \to a} f(x)[/imath] we never allow [imath]x=a[/imath].
Rather we consider [imath]x\approx a[/imath] [i.e. x close to a, but not equal to a]
Hence, if [imath]f(x)=3[/imath] then [imath]\mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h} = \mathop {\lim }\limits_{h \to 0} \dfrac{{3 - 3}}{h} = 0[~h\ne 0][/imath]



[imath][/imath][imath][/imath]
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I just wanted the student to say that the fraction will never be 0/0 while the numerator will be 0. I hate when students, for example, cross out h/h thinking it is 1.
 
Steven G said:
I just wanted the student to say that the fraction will never be 0/0 while the numerator will be 0. I hate when students, for example, cross out h/h thinking it is 1.
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But h/h is 1! To be specific, while prepare to take the limit, h is never 0, so there is nothing wrong with doing that.

What they must not do is to cross things out without thinking about that fact.

And what is being done wrong in this instance is replacing h with 0, obtaining 0/0, and continuing as if that made sense, without distinguishing between evaluation and limit. They need to back up, see that the expression equals 0 for any non-zero h, before taking the limit, and then proceed with the limit. So your first sentence is correct.
 
Dr.Peterson said:
But h/h is 1! To be specific, while prepare to take the limit, h is never 0, so there is nothing wrong with doing that.
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Completely agree.

Dr.Peterson said:
And what is being done wrong in this instance is replacing h with 0, obtaining 0/0, and continuing as if that made sense, without distinguishing between evaluation and limit. They need to back up, see that the expression equals 0 for any non-zero h, before taking the limit, and then proceed with the limit.
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Note that the OP never actually wrote 0/0. @Mathiscool0077 post#5 was correct, I think :thumbup:

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I think that Steven's original intent in post#7 was to question the OP about why the limit of 0/h (as h goes to 0) is valid and equal to 0. In other words, I think Steven wanted them to state that h never actually becomes 0.

However, it seems that Steven got a bit confused while he was explaining himself in post#10 :ROFLMAO: Perhaps it was a long day at the office?
 
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