\(\tiny\displaystyle \lim_{x\to1}\left ( \frac{x^4+1}{x^2+1} \right )+ \displaystyle \lim_{x\to1}\left ( \frac{x^4-1}{x^2+1} \right )+ \displaystyle \lim_{x\to1}\left ( \frac{x^4-1}{x^2-1} \right )\)

is gelijk aan 		A.   1
B.   2
C.   3
D.   4
E.   5
A    B    C    D    E

[ 5-7510 - op net sinds 9.12.16-()-25.8.2025 ]

Translation in   E N G L I S H

\(\displaystyle \lim_{x\to1}\left ( \frac{x^4+1}{x^2+1} \right )+ \displaystyle \lim_{x\to1}\left ( \frac{x^4-1}{x^2+1} \right )+ \displaystyle \lim_{x\to1}\left ( \frac{x^4-1}{x^2-1} \right )\)

is equal to 		A.   1
B.   2
C.   3
D.   4
E.   5

Oplossing - Solution

\(\displaystyle \lim_{x\to1}\left ( \frac{x^4+1}{x^2+1} \right )+ \displaystyle \lim_{x\to1}\left ( \frac{x^4-1}{x^2+1} \right )+ \displaystyle \lim_{x\to1}\left ( \frac{x^4-1}{x^2-1} \right ) = L\)
1ste manier :
\( L=\frac{1+1}{1+1}+\frac{1-0}{1+1}+\displaystyle \lim_{x\to 1}\frac{(x^2-1)(x^2+1)}{x^2-1} =1+0+(1^2+1)\)
2de manier : met de regel van de L'Hospital in de 2de limiet
\(L= \displaystyle \lim_{x\to 1}\frac{2x^4}{x^2+1}+\displaystyle \lim_{x\to 1}\frac{4x^3}{2x}=\frac 22 + 2 = 3\)
gricha