Please help me solve these limits... So, I need you to solve one of these limits for me, so I can see how it's done, so I can do the rest myself.

 A: OK, one. Here comes:
$$
\frac{\sqrt{u}-\sqrt{v}}{\sqrt[3]{u}-\sqrt[3]{v}}
=
\frac{u-v}{\sqrt{u}+\sqrt{v}}\cdot\frac{\sqrt[3]{u^2}+\sqrt[3]{uv}+\sqrt[3]{v^2}}{u-v}=
\frac{\sqrt[3]{u^2}+\sqrt[3]{uv}+\sqrt[3]{v^2}}{\sqrt{u}+\sqrt{v}}\underset{u,v\to1}{\longrightarrow}\frac{1+1+1}{1+1}.
$$
Equivalently, when $x\to0$, $(1\pm x)^a=1\pm ax+o(x)$ hence
$$
\frac{(1+x)^a-(1-x)^a}{(1+x)^b-(1-x)^b}=\frac{1+ax-(1-ax)+o(x)}{1+bx-(1-bx)+o(x)}=\frac{2ax+o(x)}{2bx+o(x)}\underset{x\to0}{\longrightarrow}\frac{a}b.
$$
A: You could multiply by the conjugate over the conjugate (or the appropriate generalization for the nth roots), and do some asymptotics.
eg: $\lim\limits_{x\rightarrow\infty} (x- \sqrt{x^2-5x} * \frac{x+\sqrt{x^2-5x}}{x+\sqrt{x^2-5x}})$
$= \lim\limits_{x\rightarrow\infty} \frac{x^2 - x^2+5x}{x+\sqrt{x^2-5x}}$
$= \lim\limits_{x\rightarrow\infty} \frac{5x}{x+\sqrt{x^2-5x}}$
At this point, notice that 
$\lim\limits_{x\rightarrow\infty} x+\sqrt{x^2-5x} =\lim\limits_{x\rightarrow\infty} 2x $
and substitute, 
to get $5/2$ as the answer.
