Question

dado g(x): √3x-1,encontre
lim  g(x+h)-g(x)/h
h>0

Answer 1

$g(x)=\sqrt{3x-1}\\\\g(x+h)=\sqrt{3(x+h)-1}$
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$\boxed{ \lim_{h \to 0} \frac{ \sqrt{3(x+h)-1} - \sqrt{3x-1} }{h} }$

agora temos que dar um jeito de eliminar esse h do denominador

multiplicando pelo conjulgado temos
$\frac{ \sqrt{3(x+h)-1} - \sqrt{3x-1}*(\sqrt{3(x+h)-1}+ \sqrt{3x-1}) }{h*(\sqrt{3(x+h)-1}+ \sqrt{3x-1} )} }$

resolvendo essa multiplicação no numerador temos
$(\sqrt{3(x+h)-1}* \sqrt{3(x+h)-1})+( \sqrt{3(x+h)-1}* \sqrt{3x-1})\\\\+( - \sqrt{3x-1} * \sqrt{3(x+h)-1})+(- \sqrt{3x-1}* \sqrt{3x-1})$

fazendo o calculo
$(\sqrt{3(x+h)-1})^2\\\\( \sqrt{3(x+h)-1}* \sqrt{3x-1})\\\\( - \sqrt{3x-1} * \sqrt{3(x+h)-1})\\\\(- \sqrt{3x-1})^2$

somando 
$\sqrt{3(x+h)-1}* \sqrt{3x-1})+( - \sqrt{3x-1} * \sqrt{3(x+h)-1})=0$

agora temos 
$(\sqrt{3(x+h)-1})^2+(- \sqrt{3x-1})^2\\\\3(x+h)-1+( -(3x-1))\\\\3x+3h-1-3x+1\\\\3h$

esse é o numerador
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temos a expressão
$\frac{3h}{h*(\sqrt{3(x+h)-1}+ \sqrt{3x-1} )} = \frac{3}{( \sqrt{3(x+h)-1)} + (\sqrt{3x} -1)}$

agora ja podemos calcular o limite com H tendendo a 0

$\lim_{h \to 0} \frac{3}{( \sqrt{3(x+h)-1)} + (\sqrt{3x} -1)}\\\\=\frac{3}{( \sqrt{3(x+0)-1)} + (\sqrt{3x} -1)}\\\\=\frac{3}{( \sqrt{3x-1)} + (\sqrt{3x} -1)}\\\\\\\\\\ \boxed{\boxed{\lim_{h \to 0}.. \frac{3}{2 \sqrt{3x-1} } }}$