Répondre :
f(x)=√(x+1)-√(x-1)
=(√(x+1)-√(x-1))(√(x+1)+√(x-1))/(√(x+1)+√(x-1))
=(x+1-x+1)/(√(x+1)+√(x-1))
=2/(√(x+1)+√(x-1))
donc lim(f(x), inf)=0
g(x)=(2x+sinx)/x
-1<sinx<1
donc (2x-1)/x<g(x)<(2x+1)/x
donc 2-1/x<g(x)<2+1/x
donc lim(g(x),inf)=2
=(√(x+1)-√(x-1))(√(x+1)+√(x-1))/(√(x+1)+√(x-1))
=(x+1-x+1)/(√(x+1)+√(x-1))
=2/(√(x+1)+√(x-1))
donc lim(f(x), inf)=0
g(x)=(2x+sinx)/x
-1<sinx<1
donc (2x-1)/x<g(x)<(2x+1)/x
donc 2-1/x<g(x)<2+1/x
donc lim(g(x),inf)=2