\(\displaystyle\lim_{x \to 2} \frac {2x^3-7x^2+4x+4} {x^4-8x^2+16}\)
is gelijk aan
|
A. \(\frac5{11}\) |
|---|
| B. \(\frac5{16}\) |
| C. \(\frac58\) |
| D. \(\frac54\) |
| E. 0 |
[ 5-5561 - op net sinds 30.5.13-(E)-19.10.2026 ]
Translation in E N G L I S H
\(\displaystyle\lim_{x \to 2} \frac {2x^3-7x^2+4x+4} {x^4-8x^2+16}\)
is equal to
|
A. \(\frac5{11}\) |
|---|
| B. \(\frac5{16}\) |
| C. \(\frac58\) |
| D. \(\frac54\) |
| E. 0 |
Oplossing - Solution
1ste manier : zonder de regel van de l'Hospital
\(\displaystyle\lim_{x\to 2}\frac{2x^3-7x^2+4x+4}{x^4-8x^2+16}=\displaystyle\lim_{x\to 2}\frac{\left(x-2\right)\left(2x^2-3x-2\right)}{(x^2-4)^2}=\displaystyle\lim_{x\to 2}\frac{2x^2-3x-2}{(x+2)^2(x-2)}\\=\displaystyle\lim_{x\to 2}\frac{\left(x-2\right)\left(2x+1\right)}{(x+2)^2(x-2)}=\displaystyle\lim_{x\to 2}\frac{2x+1}{(x+2)^2}=\frac{5}{16}\)
2de manier : met de regel van de l'Hospital
\(\displaystyle\lim_{x\to 2}\frac{2x^3-7x^2+4x+4}{x^4-8x^2+16}\left(=\frac{0}{0}\right)\overset{H}{=}\displaystyle\lim_{x\to 2}\frac{6x^2-14x+4}{4x^3-16x}\left(=\frac{0}{0}\right)\\\overset{H}{=}\displaystyle\lim_{x\to 2}\frac{12x-14}{12x^2-16}=\frac{24-14}{48-16}=\frac{10}{32}=\frac{5}{16}\)
