Respuesta :

x->0+ means x approach 0 from +ve side so x is ~= 0.00000000001

plug it into the eqn

sin(0.00000000001)*ln(0.00000000001)

=-2.5*10^-10

~= 0

Let f(x)=sinx;

f'(x)=cosx

Let g(x)=1/lnx;

g'(x)=-1/x(lnx)^2


as x approaches 0,

lim f(0)=sin(0)=0

lim f'(0)=cos(0)=1

lim g(0)=1/ln(0)=0

lim g'(0)=-∞


The question lim x->0+ (sin(x)ln(x)) = lim f(0)/g(0) = lim 0/0

by L'Hospital rule, lim f(x)/g(x) =  lim f'(x)/g'(x)

The limit is equal to lim x->0+ cos(0)/((1/x)*lnx^2)) = lim 1/∞ = 0


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