Wiskunde Multiple Choice Vraag

$\displaystyle\lim_{x\to\,0}\frac{\sqrt{2+\sqrt{1+x^4}}-\sqrt3}{x^4}=$	A. 1
	B. $\frac18$
	C. $\frac{\sqrt3}{4}$
	D. $\frac{\sqrt2}{10}$
	E. $\frac{\sqrt3}{12}$

$\inline\\\displaystyle\lim_{x\to\,0}\frac{\sqrt{2+\sqrt{1+x^4}}-\sqrt3}{x^4}\left(=\frac00\right)\overset{H}{=}\lim_{x\to\,0}\frac{(\sqrt{2+\sqrt{1+x^4}}-\sqrt3)(\sqrt{2+\sqrt{1+x^4}}+\sqrt3)}{x^4(\sqrt{2+\sqrt{1+x^4}}+\sqrt3)}\\\lim_{x\to\,0}\frac{2+\sqrt{1+x^4}-3}{x^4(\sqrt{2+\sqrt{1+x^4}}+\sqrt3)}=\lim_{x\to\,0}\frac{\sqrt{1+x^4}-1}{x^4(\sqrt{2+\sqrt{1+x^4}}+\sqrt3)}\left(=\frac00\right)\\\overset{H}{=}\lim_{x\to\,0}\frac{(\sqrt{1+x^4}-1)(\sqrt{1+x^4}+1)}{x^4(\sqrt{2+\sqrt{1+x^4}}+\sqrt3).(\sqrt{1+x^4}+1)}=\lim_{x\to\,0}\frac{1+x^4-1}{x^4(\sqrt{2+\sqrt{1+x^4}}+\sqrt3).(\sqrt{1+x^4}+1)}\\\lim_{x\to\,0}\frac{1}{(\sqrt{2+\sqrt{1+x^4}}+\sqrt3).(\sqrt{1+x^4}+1)}=\frac1{(\sqrt3+\sqrt3)(\sqrt1+1)}=\frac{1}{4\sqrt3}=\frac{\sqrt3}{12}\quad(\approx 0,144)$

^{gricha - vA130 - 17.2.2026}

$\displaystyle\lim_{x\to\,0}\frac{\sqrt{2+\sqrt{1+x^4}}-\sqrt3}{x^4}=$	A. 1
	B. \(\frac18\)
	C. \(\frac{\sqrt3}{4}\)
	D. \(\frac{\sqrt2}{10}\)
	E. \(\frac{\sqrt3}{12}\)

$\displaystyle\lim_{x\to\,0}\frac{\sqrt{2+\sqrt{1+x^4}}-\sqrt3}{x^4}=$	A. 1
	B. \(\frac18\)
	C. \(\frac{\sqrt3}{4}\)
	D. \(\frac{\sqrt2}{10}\)
	E. \(\frac{\sqrt3}{12}\)