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Derivation af oa product

  1. Nov 13, 2011 #1
    What do I do wrong here:

    [tex]\frac{f(x+h)g(x+h)-f(x)g(x)}{x+h-x}=\frac{f(x+h)g(x+h)}{h}-\frac{f(x)g(x)}{h}=\frac{f(x+h)}{h}g(x+h)-\frac{f(x)}{h}g(x)=\lim_{h \to 0}\frac{f(x+h)}{h}\lim_{h \to 0}g(x+h)-\lim_{h \to 0}\frac{f(x)}{h}g(x)[/tex]

    [tex]\lim_{h \to 0}g(x+h)=g(x)[/tex]

    [tex]\lim_{h \to 0}\frac{f(x+h)}{h}\lim_{h \to 0}g(x+h)-\lim_{h \to 0}\frac{f(x)}{h}g(x)=\lim_{h \to 0}\frac{f(x+h)}{h}g(x)-\lim_{h \to 0}\frac{f(x)}{h}g(x)=(\lim_{h \to 0}\frac{f(x+h)}{h}-\lim_{h \to 0}\frac{f(x)}{h})g(x)=(\lim_{h \to 0}\frac{f(x+h)-f(x)}{h})g(x)=\frac{df}{dx}g(x)[/tex]

    I know how they derieve the derivation of a product:


    But how come what I did above does not work?
    Last edited: Nov 13, 2011
  2. jcsd
  3. Nov 13, 2011 #2
    You did something like

    [tex]\lim_{x\rightarrow a}{f(x)g(x)+h(x)}=\lim_{x\rightarrow a}{f(x)}\lim_{x\rightarrow a}{g(x)}+\lim_{x\rightarrow a}{h(x)}[/tex]

    But this is not true. You can't do that.

    It is ONLY true if [itex]\lim_{x\rightarrow a}{f(x)}[/itex], [itex]\lim_{x\rightarrow a}{g(x)}[/itex] AND [itex]\lim_{x\rightarrow a}{h(x)}[/itex] converge. In your example, you don't have that. For example

    [tex]\lim_{h\rightarrow 0}{\frac{f(x)}{h}}[/tex]

    does not converge.
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