Dadas as funções f(x) = x2, f(x) = 2x2+3 e f(x) = 1/x, determine, para cada uma:
lim [ f(2 + z) – f(2) ] / z “.
z->0
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lim [ f(2 + z) – f(2) ] / z
z->0
f(x)=x²
lim [ (2+z)² – 2² ] / z
z->0
lim [2z+z² ] / z
z->0
lim [4+2z+z² – 4 ] / z
z->0
lim 2+z =2+0 =2
z->0
================================================
f(x) = 2x² +3
lim [ f(2 + z) – f(2) ] / z
z->0
lim [2*(2+z)+3– 2*2²+3] / z
z->0
lim [2*(2+z)+3– 2*2²+3] / z
z->0
lim [4+2z+3– 4+3] / z
z->0
lim [2z] / z = lim 2 =2
z->0 z-->0
#########################################
f(x) = 1/x
lim [ f(2 + z) – f(2) ] / z
z->0
lim [ 1/(2+z) – 1/2 ] / z
z->0
lim [ (2-(2+z))/2(2+z) ] / z
z->0
lim [ z/2(2+z) ] / z
z->0
lim 1/2(2+z) =1/(4+0)=1/4
z->0
z->0
f(x)=x²
lim [ (2+z)² – 2² ] / z
z->0
lim [2z+z² ] / z
z->0
lim [4+2z+z² – 4 ] / z
z->0
lim 2+z =2+0 =2
z->0
================================================
f(x) = 2x² +3
lim [ f(2 + z) – f(2) ] / z
z->0
lim [2*(2+z)+3– 2*2²+3] / z
z->0
lim [2*(2+z)+3– 2*2²+3] / z
z->0
lim [4+2z+3– 4+3] / z
z->0
lim [2z] / z = lim 2 =2
z->0 z-->0
#########################################
f(x) = 1/x
lim [ f(2 + z) – f(2) ] / z
z->0
lim [ 1/(2+z) – 1/2 ] / z
z->0
lim [ (2-(2+z))/2(2+z) ] / z
z->0
lim [ z/2(2+z) ] / z
z->0
lim 1/2(2+z) =1/(4+0)=1/4
z->0
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