I need help solving these limits I've been struggling to solve the following limits for about an hour. I've tried using conjugates as well as common factors, but it gets me nowhere. Wolfram Alpha does not provide steps for these limits. I could really use some help.
 (x greater than 1) (SOLVED)

 (SOLVED)
Thanks for taking your time!
 A: For question 1: You can factorize $x^2+2x-3 = (x+3)(x-1)$ to get $\frac{1}{\sqrt{x-1}}(\frac{1}{2}-\frac{1}{\sqrt{x+3}})$. After that, I'd suggest calculating Taylor expansions at x=1 for $\frac{1}{\sqrt{x-1}}$ and $\frac{1}{2}-\frac{1}{\sqrt{x+3}}$ (you can do this with Wolfram Alpha). Btw, the limit is zero, I think (used a calculator entering x slightly >1).
I'd post this as a comment, if possible, but I don't have enough reputation, so sorry that the answer is incomplete :(
EDIT:
For the second question, I have this idea:
First (inspired from André Nicolas' comment), we multiply by $\frac{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}$, so that we have
$$\frac{x+\sqrt{x+\sqrt{x}}-x}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}=\frac{\sqrt{x+\sqrt{x}}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}$$
Next, we devide numerator and denominator by $\sqrt{x}$:
$$=\frac{\sqrt{\frac{x+\sqrt{x}}{x}}}{\sqrt{\frac{x+\sqrt{x+\sqrt{x}}}{x}}+1}
=\frac{\sqrt{1+\frac{1}{\sqrt{x}}}}{\sqrt{1+\sqrt{\frac{x+\sqrt{x}}{x^2}}}+1}
=\frac{\sqrt{1+\frac{1}{\sqrt{x}}}}{\sqrt{1+\sqrt{\frac{1}{x}+\frac{1}{\sqrt{x}^3}}}+1}$$
Now we can see that $\frac{1}{x}$, $\frac{1}{\sqrt{x}}$ and $\frac{1}{\sqrt{x}^3}$ will all become zero for $x\rightarrow\infty$, so what remains is $$\frac{\sqrt{1}}{\sqrt{1}+1}=\frac{1}{1+1} = 0.5$$
A: Addressing the third limit here. 
Substitution yields:
$$\frac{\sqrt[]{1-(-8)}-3}{2+(-8)^{2/3}}=\frac{3-3}{2-2}=\frac{0}{0}$$
which is an indeterminate form. Therefore we can apply L'Hospital's rule to obtain 
$$\lim_{x\rightarrow-8}{\frac{\sqrt[]{1-x}-3}{2+\sqrt[3]{x}}}=\lim_{x\rightarrow-8}{\frac{-\frac{1}{2\sqrt[]{1-x}}}{\frac{1}{3x^{2/3}}}}=-\frac{3}{2}\lim_{x\rightarrow-8}{\frac{x^{2/3}}{\sqrt[]{1-x}}}=-\frac{3}{2}\cdot\frac{4}{3}=-2$$
$\textbf{Edit:}$
Another solution without L'Hospital's rule:
First multiply by the conjugate of the numerator over itself:
$$\lim_{x\rightarrow-8}{\frac{\sqrt[]{1-x}-3}{2+\sqrt[3]{x}}}=\lim_{x\rightarrow-8}{\frac{\sqrt[]{1-x}-3}{2+\sqrt[3]{x}}}\frac{\sqrt[]{1-x}+3}{\sqrt[]{1-x}+3}=\lim_{x\rightarrow-8}{\frac{-(x+8)}{(2+\sqrt[3]{x})(\sqrt[]{1-x}+3)}}$$
Now to avoid the indeterminate form we want to find a way to get a $x+8$ in the denominator, to cancel the one in the numerator. Multiplying the top and bottom by $4-2\sqrt[3]{x}+x^{2/3}$ will do the trick. That's because of the sum of cubes formula: $a^3+b^3=(a+b)(a^2-ab+b^2)$.
So from above
$$\lim_{x\rightarrow-8}{\frac{\sqrt[]{1-x}-3}{2+\sqrt[3]{x}}}=\lim_{x\rightarrow-8}{\frac{-(x+8)}{(2+\sqrt[3]{x})(\sqrt[]{1-x}+3)}}\cdot\frac{4-2\sqrt[3]{x}+x^{2/3}}{4-2\sqrt[3]{x}+x^{2/3}}=\lim_{x\rightarrow-8}{\frac{-(x+8)(4-2\sqrt[3]{x}+x^{2/3})}{(2+\sqrt[3]{x})(4-2\sqrt[3]{x}+x^{2/3})(\sqrt[]{1-x}+3)}}=\lim_{x\rightarrow-8}{\frac{-(x+8)(4-2\sqrt[3]{x}+x^{2/3})}{(x+8)(\sqrt[]{1-x}+3)}}$$
Canceling the $x+8$ gives
$$\lim_{x\rightarrow-8}{\frac{\sqrt[]{1-x}-3}{2+\sqrt[3]{x}}}=\lim_{x\rightarrow-8}{\frac{-(4-2\sqrt[3]{x}+x^{2/3})}{\sqrt[]{1-x}+3}}=-\frac{4+4+4}{3+3}=-\frac{12}{6}=-2$$
