Limit $\lim _ { x \to - \infty } \frac{ x }{(\sqrt{ x ^2} + \sqrt{ x ^4 + 1} ) + \sqrt{ 2} ) ( \sqrt{ x ^4 + 1} + x ^2) }$

Question

When I try to direct substitute the value I am getting $\infty \cdot 0$. Actually it's not the entire question it is the last step where I reached and I don't know how to proceed further. The actual question is

$$\lim_{ x \to \pm \infty } x *( \sqrt{ x^2 + \sqrt{ x^4 + 1} } - x \sqrt{2})$$

Answer 1

For $x\to +\infty$

$$x \cdot\left( \sqrt{ x^2 + \sqrt{ x^4 + 1} } - x \sqrt{2}\right)=x \cdot\left( \sqrt{ x^2 + \sqrt{ x^4 + 1} } - x \sqrt{2}\right)\frac{\sqrt{ x^2 + \sqrt{ x^4 + 1} } + x \sqrt{2}}{\sqrt{ x^2 + \sqrt{ x^4 + 1} } + x \sqrt{2}}=\\=x\cdot\frac{x^2 + \sqrt{ x^4 + 1} - 2x^2}{\sqrt{ x^2 + \sqrt{ x^4 + 1} } + x \sqrt{2}}=x\cdot\frac{\sqrt{ x^4 + 1}-x^2}{\sqrt{ x^2 + \sqrt{ x^4 + 1} } + x \sqrt{2}}\frac{\sqrt{ x^4 + 1}+x^2}{\sqrt{ x^4 + 1}+x^2}=\\=\frac{x}{\left(\sqrt{ x^2 + \sqrt{ x^4 + 1} } + x \sqrt{2}\right)\left(\sqrt{ x^4 + 1}+x^2\right)}=\\=\frac{1}{\left(\sqrt{ 1 + \sqrt{ 1 + 1/x^4} } + \sqrt{2}\right)\left(\sqrt{ x^4 + 1}+x^2\right)}\to 0$$

For $x\to -\infty$

$$x \cdot\left( \sqrt{ x^2 + \sqrt{ x^4 + 1} } - x \sqrt{2}\right)(\to -\infty \cdot \infty)\to-\infty$$

Answer 2

And this is my inexperienced approach of the problem.

First you can extract a $x$ infront of the parentheses. Let the expression of which we seek the limit be A. Then $$A = x(\sqrt{x^2 + \sqrt{x^4 + 1}} - \sqrt{2}x) = x^2(\sqrt{1 + \sqrt{1 + \frac{1}{x^4}}} - \sqrt{2})$$

So let's multiply by $\frac{\sqrt{1 + \sqrt{1 + \frac{1}{x^4}}} + \sqrt{2}}{\sqrt{1 + \sqrt{1 + \frac{1}{x^4}}} + \sqrt{2}}$:

$$A = \frac{x^2\big(\sqrt{1 + \sqrt{1 + \frac{1}{x^4}}} - \sqrt{2}\big)\big(\sqrt{1 + \sqrt{1 + \frac{1}{x^4}}} + \sqrt{2}\big)}{\sqrt{1 + \sqrt{1 + \frac{1}{x^4}}} + \sqrt{2}} = \frac{x^2\big( \sqrt{1 + \frac{1}{x^4}} - 1 \big)}{\sqrt{1 + \sqrt{1 + \frac{1}{x^4}}} + \sqrt{2}}$$

We'll repeat the analogue action by multiplying by $\frac{\sqrt{1 + \frac{1}{x^4}} + 1 }{\sqrt{1 + \frac{1}{x^4}} + 1 }$ so that we get:

$$A = \frac{x^2\big(\sqrt{1 + \frac{1}{x^4}} - 1 \big)\big(\sqrt{1 + \frac{1}{x^4}} + 1 \big)}{\big(\sqrt{1 + \sqrt{1 + \frac{1}{x^4}}} + \sqrt{2}\big)\big( \sqrt{1 + \frac{1}{x^4}} + 1 \big)} = \frac{x^2.\frac{1}{x^4}}{\big(\sqrt{1 + \sqrt{1 + \frac{1}{x^4}}} + \sqrt{2}\big)\big( \sqrt{1 + \frac{1}{x^4}} + 1 \big)}$$ So at the end: $$A = \frac{1}{x^2\big(\sqrt{1 + \sqrt{1 + \frac{1}{x^4}}} + \sqrt{2}\big)\big( \sqrt{1 + \frac{1}{x^4}} + 1 \big)}$$

And $$lim_{n \to \pm \infty}A = \frac{1}{\infty} = 0$$

Answer 3

Alternatively, for $x>0$: $$\color{red}{0<}\sqrt{x^2+\sqrt{x^4+1}}-x\sqrt{2} \iff 2x^2<x^2+\sqrt{x^4+1} \iff 0<1;\\ \sqrt{x^2+\sqrt{x^4+1}}-x\sqrt{2}\color{red}{<\frac 1{x^2}} \iff \sqrt{x^2+\sqrt{x^4+1}}<\frac 1{x^2}+x\sqrt{2} \iff \\ x^2+\sqrt{x^4+1}<\frac 1{x^4}+\frac{2\sqrt{2}}{x}+2x^2 \iff x^4\sqrt{x^4+1}<x^6+2\sqrt{2}x^3+1 \iff \\ x^4\sqrt{x^4+1}<x^4\sqrt{\left(x^2+\frac 1{x^2}\right)^2}=x^6+x^2 <x^6+2\sqrt{2}x^3+1 \iff \\ x^2<x^3<2\sqrt{2}x^3+1.$$

Hence: $$0\le \lim_{ x \to + \infty } x *( \sqrt{ x^2 + \sqrt{ x^4 + 1} } - x \sqrt{2})\le \lim_{ x \to + \infty } x * \frac 1{x^2}=\lim_{ x \to + \infty } \frac1x =0.$$