This one is giving me some truble to modify correctly to allow me get $$\lim\limits_{x\to+\infty} \frac{(x + (x +(x)^{\frac 12})^{\frac 12})^{\frac 12}}{(x+1)^{\frac 12}} = 1$$
I keep on trying fatoring and turning it around but i can't seem to get away from $0/0$
(i gotta admit, i'm pretty bad at limits, i would welcome any recommendations for some good site for learning limits and possibly some solved problems to get experience on)
On
The method is always the same, spot the dominant factor, divide everthing else by it (so this everything else goes to $0$) and either the limit appears naturally either you can now perform taylor expansions with things that are $\ll 1$.
This is what lab bhattacharjee did in his answer.
But notice also that you can spot the result without going too formal with the $o(\cdot)$ and expliciting the infinitesimal quantities, by using equivalents when there is no doubt about the dominance relation.
In our case notice that $\displaystyle{\frac{\sqrt x}{x}=\frac{1}{\sqrt x}\to 0}$ when $x\to+\infty$.
This means that $\sqrt x=o(x)$ but more importantly that $(x+\sqrt x)\sim x$
So in the expression you gave, we can have a chain reaction
$\sqrt{x+\sqrt{\vphantom{|} x+\sqrt{\vphantom{|}x}}}\sim\sqrt{\vphantom{|}x+\sqrt{\vphantom{|} x}}\sim\sqrt{x}$
For the denominator $\sqrt{\vphantom{|}x+1}\sim\sqrt x$
And finally you get $f(x)\sim\frac{\sqrt x}{\sqrt x}\sim1$ which is the searched limit.
On
Squeezing seems the most natural and straightforward approach to me: $$\sqrt{x}\leq \sqrt{x+\sqrt{x+\sqrt{x}}} \leq \sqrt{x+\sqrt{2x}}\leq\sqrt{x}\left(1+\frac{1}{\sqrt{2x}}\right)$$ and both $\lim_{x\to +\infty}\frac{\sqrt{x}}{\sqrt{x+1}}$ and $\lim_{x\to +\infty}\frac{\sqrt{x}}{\sqrt{x+1}}\left(1+\frac{1}{\sqrt{2x}}\right)$ clearly equal $1$.
HINT:
$$\dfrac{\sqrt{x+\sqrt{x+\sqrt x}}}{\sqrt{x+1}}=\dfrac{\sqrt{1+\sqrt{\dfrac1x+\sqrt{\dfrac1{x^3}}}}}{\sqrt{1+\dfrac1x}}$$