Answered

Etudier les limites suivantes:
[tex] \lim_{x \to \infty} \sqrt{x+1}- \sqrt{x-1} [/tex]
[tex] \lim_{x \to \infty} \frac{2x+sinx}{x} [/tex]

Sagot :

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