SchacliverraSo
SchacliverraSo SchacliverraSo
  • 13-10-2016
  • Mathematics
contestada

im h--> 0

((Sqrt 9+h)-3)/h

Respuesta :

HomertheGenius
HomertheGenius HomertheGenius
  • 25-10-2016
[tex] \lim_{h \to 0} \frac{ \sqrt{9+h}-3}{h} = \frac{ \sqrt{9+0} -3}{0}= \frac{0}{0} \\ \lim_{h \to 0} \frac{ \sqrt{9+h}-3}{h}* \frac{ \sqrt{9+h}+3 }{ \sqrt{9+h} +3}= \\ = \lim_{h \to 0} \frac{9+h-9}{h( \sqrt{9+h} +3)} = \\ \lim_{n \to 0} \frac{h}{h( \sqrt{9+h}+3) }= \\ = \frac{1}{ \sqrt{9} +3} [/tex]= 
= 1/6
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