Encuentre el límite $\lim\limits _{x\rightarrow \infty }\left(\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}} -\sqrt{2x^{4}}\right)$

Question

$\displaystyle \lim\limits _{x\rightarrow \infty }\left(\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}}\displaystyle -\sqrt{2x^{4}}\right)$

$\displaystyle \lim\limits _{x\rightarrow \infty }\left(\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}} -\sqrt{2x^{4}}\right)$$\displaystyle =\displaystyle \lim\limits _{x\rightarrow \infty }\dfrac{\left(\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}} -\sqrt{2x^{4}}\right)\left(\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}} +\sqrt{2x^{4} \ }\right)}{\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}} +\sqrt{2x^{4} \ }} =$

$\displaystyle =\lim\limits _{x\rightarrow \infty }\dfrac{x^{2}\sqrt{x^{4} +1} -x^{4}}{\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}} +\sqrt{2x^{4} \ }}$

¿Cuál debería ser el siguiente paso? ¡Por favor ayuda!

Answer 1

Ahora, usa eso $$\sqrt{x^4+1}-x^2=\frac{1}{\sqrt{x^4+1}+x^2}.$$

Answer 2

Use el truco dos veces para el numerador para obtener

$$\dfrac{x^{2}\sqrt{x^{4} +1} -x^{4}}{\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}} +\sqrt{2x^{4} \ }}=\dfrac{x^{2}\sqrt{x^{4} +1} -x^{4}}{\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}} +\sqrt{2x^{4} \ }}\cdot \dfrac{x^{2}\sqrt{x^{4} +1} +x^{4}}{x^{2}\sqrt{x^{4} +1} +x^{4}}=$$

$$=\dfrac{x^4}{\left(\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}} +\sqrt{2x^{4} \ }\right)\left(x^{2}\sqrt{x^{4} +1} +x^{4}\right)}\sim\frac{1}{4\sqrt 2 x^2}\to 0$$

Answer 3

$$ \sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}}-\sqrt{2x^4}= \dfrac{x^{2}\sqrt{x^{4} +1} -x^{4}}{\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}} +\sqrt{2x^{4}}}= \dfrac{x^{2}\sqrt{x^{4} +1} -x^{4}}{\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}} +\sqrt{2x^{4} \ }}\cdot\frac{x^{2}\sqrt{x^{4} +1} +x^{4}}{x^{2}\sqrt{x^{4} +1} +x^{4}}\\=\frac{x^4}{\big(\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}} +\sqrt{2x^{4}}\big)\big(x^{2}\sqrt{x^{4} +1}+x^4\big)}\\=\frac{1}{\big(\sqrt{1 +\sqrt{1+x^{-4}}} +\sqrt{2}\big)\big(\sqrt{x^{4}+1}+x^2\big)}\,\to\,0 $$

Answer 4

Otro enfoque es usar la sustitución $x=\cot t$: \begin{align} \lim_{x\to\infty}\left(\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}}-\sqrt{2x^{4}}\right) &=\lim_{t\to0}\dfrac{\sqrt{\cos^2t+\cos t}-\sqrt{2\cos^2t}}{\sin t} \\ &=\lim_{t\to0}\dfrac{\cos t(1-\cos t)}{\sin t(\sqrt{\cos^2t+\cos t}+\sqrt{2\cos^2t})} \\ &=\lim_{t\to0}\dfrac{\cos t}{\sqrt{\cos^2t+\cos t}+\sqrt{2\cos^2t}}\dfrac{2\sin^2\frac{t}{2}}{\frac{t^2}{2}}\dfrac{t}{\sin t}\dfrac{t}{2} \\ &=\color{blue}{0} \end{align}