De limiet

\(\small\displaystyle \lim_{x \to +\infty}\: \left ( \sqrt{x^2+1}-\sqrt{x^3+1} \right )\)
A.  =  − 1
B.  =  minoneindig
C.  =  plusoneindig
D.  =   0
E.  bestaat niet
A    B    C    D    E

[ 5-6960 - op net sinds 20.10.2020-(E)-19.10.2026 ]

Translation in   E N G L I S H

The limit
\(\displaystyle \lim_{x \to \infty}\: \left ( \sqrt{x^2+1}-\sqrt{x^3+1} \right )\)
A.   = 1
B.   = − ∞
C.   = ∞
D.   = 0
E.   does not exist

Oplossing - Solution

  \(\displaystyle \lim_{x \to \infty}\: \left ( \sqrt{x^2+1}-\sqrt{x^3+1} \right )\) \(\displaystyle=\lim_{x\to+\infty}\frac{(\sqrt{x^2+1}-\sqrt{x^3+1})\cdot(\sqrt{x^2+1}+\sqrt{x^3+1})}{\sqrt{x^2+1}+\sqrt{x^3+1}}\)
\(\displaystyle =\lim_{x\to+\infty}\frac{x^2+1-(x^3+1)}{\sqrt{x^2+1}+\sqrt{x^3+1}}=\lim_{x\to+\infty}\frac{x^3\left(\frac{1}{x}-1\right)}{x^2\sqrt{\frac{1}{x^2}+\frac{1}{x^4}}+x^2\sqrt{\frac{1}{x}+\frac{1}{x^4}}}\)
\(\displaystyle=\lim_{x\to+\infty}\frac{x\left(\frac{1}{x}-1\right)}{\sqrt{\frac{1}{x^2}+\frac{1}{x^4}}+\sqrt{\frac{1}{x}+\frac{1}{x^4}}}=\frac{+\infty\cdot(-1)}{+0+0}=\frac{-\infty}{+0}=-\infty\)
( sommigen zullen i.p.v. + 0 + 0 liever schrijven  0+ + 0+ )
gricha