Limits without L'Hopitals Rule ( as I calculate it?) Prove that:
$\lim z \to \infty \left (z^2 +\sqrt{z^{4}+2z^{3}}-2\sqrt{z^{4}+z^{3}}\right )=\frac{-1}{4}$

Let $\displaystyle z = \frac{1}{y}\;,$ So when $z\rightarrow \infty\;,$ Then $y\rightarrow 0$
So limit convert into $$\displaystyle \lim_{y\rightarrow 0}\frac{1+\sqrt{1+2y}-2\sqrt{1+y}}{y^2} = \lim_{y\rightarrow 0}\frac{1+(1+2y)^{\frac{1}{2}}-2(1+y)^{\frac{1}{2}}}{y^2}$$
Now Using $$\displaystyle \bullet (1+t)^n = 1+nt+\frac{n(n-1)t^2}{2}+\frac{n(n-1)(n-2)}{6}t^3+.....$$
So we get $$\displaystyle \lim_{y\rightarrow 0}\frac{1+(1+ y-\frac{y^2}{2}-\frac{y^3}{6}+...)-2(1+\frac{y}{2}-\frac{y^2}{8}-\frac{z^3}{16}....)}{y^2}$$
So we get limit $$\displaystyle = -\frac{1}{4}$$
