Limits without L'Hopitals Rule ( as I calculate it?)

Question

Prove that:

$\lim z \to \infty \left (z^2 +\sqrt{z^{4}+2z^{3}}-2\sqrt{z^{4}+z^{3}}\right )=\frac{-1}{4}$

Answer 1

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 $$\displaystyle = -\frac{1}{4}$$

Answer 2

$$\begin{align} z^2+\sqrt{z^4+2z^3}-2\sqrt{z^4+z^3} &=(\sqrt{z^4+2z^3}-\sqrt{z^4+z^3})+(z^2-\sqrt{z^4+z^3})\\ &={z^3\over\sqrt{z^4+2z^3}+\sqrt{z^4+z^3}}-{z^3\over z^2+\sqrt{z^4+z^3}}\\ &={z^3(z^2-\sqrt{z^4+2z^3})\over(\sqrt{z^4+2z^3}+\sqrt{z^4+z^3})(z^2+\sqrt{z^4+z^3})}\\ &={z^3(-2z^3)\over(\sqrt{z^4+2z^3}+\sqrt{z^4+z^3})(z^2+\sqrt{z^4+z^3})(z^2+\sqrt{z^4+2z^3})}\\ \end{align}$$

Can you take it from here?

Answer 3

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: $$\text {Firstly, }a>0\implies (1+a/2)^2=1+a+(a/2)^2>1+a\implies $$ $$1+a/2>\sqrt {1+a}.$$ $$\text {Secondly , for } a\in (0,1), \text { let } 1+a=1/(1-b) \text { with }b=a/(1+a)\in (0,1).$$ $$\text {Now, }b\in (0,1)\implies (1-b/2)^2=1-b+b^2/2>1-b\implies (1-b/2)>\sqrt {1-b}.$$ $$\text {Therefore }1+\frac {a/2}{1+a}=\frac {1}{1-b/2}<\frac {1}{\sqrt {1-b}}=\sqrt {1+a}.$$ $$ \text {Altogether we have } a\in (0,1)\implies 1+a/2>\sqrt {1+a}>1+\frac{a/2}{1+a}.$$

Answer 4

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 $$\lim_{z\to\infty}z^2 +\sqrt{z^{4}+2z^{3}}-2\sqrt{z^{4}+z^{3}}=-\frac14.$$