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We can correct your mistake only if you share your work - otherwise we would have to throw stones in dark!I know that the derivative of 1 is obviously 0, but using the first principles definition of the derivative limit rules for a constant, it is 1. Where am I going wrong here? Thanks.

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Did you even read the question? It says the derivative of 1. That is \(f(x)=1\).Isn't it that if f(x) = C, then lim x-> n = C? Then substitute C=1.

In any case \(\dfrac{f(x+h)-f(x)}{h}=\dfrac{1-1}{h}=\dfrac{0}{h}=0\)

Thus \(\mathop {\lim }\limits_{h \to 0} \dfrac{f(x+h)-f(x)}{h}=0\)

The first derivative rule for a constant isI know that the derivative of 1 is obviously 0, but using the first principles definition of the derivative limit rules for a constant, it is 1. Where am I going wrong here? Thanks.

\(\displaystyle f(x) = c \implies \lim_{h \rightarrow 0} \dfrac{f(x + h) - f(x)}{h} = \lim_{h \rightarrow 0} \dfrac{c- c}{h} =\)

\(\displaystyle \lim_{h \rightarrow 0} \dfrac{0}{h} =\lim_{h \rightarrow 0} 0 = 0.\)

You are mixing this up with a completely different rule.

\(\displaystyle f(x) = x \implies \lim_{h \rightarrow 0} \dfrac{f(x + h) - f(x)}{h} = \lim_{h \rightarrow 0} \dfrac{x + h- x}{h} =\)

\(\displaystyle \lim_{h \rightarrow 0} \dfrac{h}{h} =\lim_{h \rightarrow 0} 1 = 1.\)

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We can tell where you are going wrong when you don't show what you did!I know that the derivative of 1 is obviously 0, but using the first principles definition of the derivative limit rules for a constant, it is 1. Where am I going wrong here? Thanks.

The limit of WHAT "as z=x->n? And what is "n"? Where did that come from? It is true that \(\displaystyle \lim_{x\to x_0} f(x)= f(x_0)= C\) where \(\displaystyle x_0\) is any value of C. That says that f(x)= C isIsn't it that if f(x) = C, then lim x-> n = C?

Then substitute C=1.