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