帮忙求几个极限

都是x趋近正无穷的

1.sqr(x+sqr(x+sqr(x)))-sqr(x)(sqr就是根号）
2.(tanx)^tan2x
3.((sqr(x^2+1))/(x+1))^x
那个- -第二个是趋近π/4的……
2题我已经解决了，高手在哪里……

1
lim(x→+∞) √(x+√(x+√x))-√x
=lim(x→+∞) [(x+√(x+√x))-x]/[√(x+√(x+√x))+√x]
=lim(x→+∞) [√(x+√x)]/[√(x+√(x+√x))+√x]
=lim(x→+∞) {[√(x+√x)]/√x}/{[√(x+√(x+√x))+√x]/√x}
=lim(x→+∞) √(1+1/√x) / {√{1+√[x^(-1)+x^(-3/2)]} +1}
=lim(x→+∞) 1/{1+1}
=1/2;

2
lim(x→π/4) (tanx)^tan2x
=lim(x→π/4) [1+(tanx-1)]^tan2x
=lim(x→π/4) [1+(tanx-1)]^{[1/(tanx-1)]·(tanx-1)·tan2x}
=lim(x→π/4) e^[(tanx-1)·tan2x]
=e^{lim(x→π/4) (tanx-1)·2tanx/(1-tan^2 x)}
=e^{lim(x→π/4) (tanx-1)·2tanx/[(1-tanx)(1+tanx)}
=e^{lim(x→π/4) -2·tanx/(1+tanx)
=e^(-1)
=1/e

3
lim(x→+∞) ((√(x^2+1))/(x+1))^x
=e^{lim(x→+∞) x·ln((√(x^2+1))/(x+1))
令x=tanu,得:
=e^{lim(u→π/2) tanu·ln((secu)/(tanu+1))
=e^{lim(u→π/2) tanu·ln(1/(sinu+cosu))
=e^{lim(u→π/2) tanu·ln[1+ 1/(sinu+cosu)-1]
注意:当t→0时,ln(1+t)～t;则:
=e^{lim(u→π/2) tanu·[1/(sinu+cosu)-1] }
=e^{lim(u→π/2) (sinu -sin^2 u -sinu·cosu)/(sinu·cosu +cos^2 u) }