Limits without L'Hopitals Rule ( as I calculate it?) Prove that:
$\lim z \to \infty \left (z^2 +\sqrt{z^{4}+2z^{3}}-2\sqrt{z^{4}+z^{3}}\right )=\frac{-1}{4}$
 A: 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 $$\displaystyle = -\frac{1}{4}$$
A: $$\begin{align}
z^2+\sqrt{z^4+2z^3}-2\sqrt{z^4+z^3}
&=(\sqrt{z^4+2z^3}-\sqrt{z^4+z^3})+(z^2-\sqrt{z^4+z^3})\\
&={z^3\over\sqrt{z^4+2z^3}+\sqrt{z^4+z^3}}-{z^3\over z^2+\sqrt{z^4+z^3}}\\
&={z^3(z^2-\sqrt{z^4+2z^3})\over(\sqrt{z^4+2z^3}+\sqrt{z^4+z^3})(z^2+\sqrt{z^4+z^3})}\\
&={z^3(-2z^3)\over(\sqrt{z^4+2z^3}+\sqrt{z^4+z^3})(z^2+\sqrt{z^4+z^3})(z^2+\sqrt{z^4+2z^3})}\\
\end{align}$$
Can you take it from here?
A: HINT.Here is something useful for handling the roots without calculus or infinite series. When you have something like $\sqrt {z^4+2 z^3} $  with $z>0$ write it as $ z^2 \sqrt {1+a} $ where,in this case $a=2/z$,and as $z\to \infty $ you need to consider only $a\in (0.1)$, with $a\to 0$ as $z\to \infty. $ To get get upper and lower bounds for $\sqrt {1+a}$ when $a\in (0,1)$ we have: $$\text {Firstly, }a>0\implies (1+a/2)^2=1+a+(a/2)^2>1+a\implies $$ $$1+a/2>\sqrt {1+a}.$$  $$\text {Secondly , for  }  a\in (0,1), \text { let  } 1+a=1/(1-b) \text { with }b=a/(1+a)\in (0,1).$$ $$\text {Now, }b\in (0,1)\implies (1-b/2)^2=1-b+b^2/2>1-b\implies (1-b/2)>\sqrt {1-b}.$$ $$\text {Therefore }1+\frac {a/2}{1+a}=\frac {1}{1-b/2}<\frac {1}{\sqrt {1-b}}=\sqrt {1+a}.$$  $$ \text {Altogether we have } a\in (0,1)\implies 1+a/2>\sqrt {1+a}>1+\frac{a/2}{1+a}.$$
A: Use Taylor's formula at order $2$:
\begin{align*}
z^2 +\sqrt{z^{4}+2z^{3}}-2\sqrt{z^{4}+z^{3}}&=z^2\biggl(1+\sqrt{1+\frac2z}-2\sqrt{1+\frac1z}\biggr)\\&=z^2\biggl(1+1+\frac1z-\frac18\frac4{z ^2}+o\Bigl(\frac1{z^2}\Bigr)-2-\frac1z+2\frac1{8z^2}+o\Bigl(\frac1{z^2}\Bigr)\biggr)\\
&=z^2\biggl(-\frac1{4z^2}+o\Bigl(\frac1{z^2}\Bigr)\biggr)=-\frac14+o(1),
\end{align*}
hence $$\lim_{z\to\infty}z^2 +\sqrt{z^{4}+2z^{3}}-2\sqrt{z^{4}+z^{3}}=-\frac14.$$
