\(\displaystyle \lim_{x \to 4} \: {\frac{x^2-16}{2-\sqrt x}}\)

is gelijk aan
A.   − 32
B.   − 24
C.   − 16
D.   0
E.   8
F.   32
A   B   C   D   E   F

[ 5-6959 - op net sinds 16.1.13-(E)-17.10.2026 ]

Translation in   E N G L I S H

\(\displaystyle \lim_{x \to 4} \: {\frac{x^2-16}{2-\sqrt x}}\)

is equal to
A.  − 32
B.  − 24
C.  − 16
D.  0
E.  8
F.  32

Oplossing - Solution

1ste manier :
\(\displaystyle\lim_{x\to 4}\:{\frac{x^2-16}{2-\sqrt x}}=\displaystyle\lim_{x\to 4}\:\frac{(x-4)(x+4)}{2-\sqrt x}=\displaystyle\lim_{x\to 4}\:\frac{(\sqrt x-2)(\sqrt x+2)(x+4)}{2-\sqrt x}\\=\displaystyle\lim_{x\to 4}\:(\sqrt x+2)(x+4)=-(2+2)(4+4)=-4.8=-32\)
2de manier :
\(\displaystyle\lim_{x\to 4}\:{\frac{x^2-16}{2-\sqrt x}}=\displaystyle\lim_{x\to 4}\:\frac{(x^2-16)(2+\sqrt x)}{(2-\sqrt x)(2+\sqrt x)}=\displaystyle\lim_{x\to 4}\:\frac{(x-4)(x+4)(2+\sqrt x)}{4-x}\\=\displaystyle\lim_{x\to 4}\:(x+4)(2+\sqrt x)=-(4+4)(2+2)=-8.4=-32\)
3de manier :
\(\displaystyle\lim_{x\to 4}\:{\frac{x^2-16}{2-\sqrt x}}\left(=\frac00\right)\overset{H}{=}\displaystyle\lim_{x\to 4}\:\frac{2x-0}{0-\frac{1}{2\sqrt x}}=\displaystyle\lim_{x\to 4}\:(-4x\sqrt x)=-4.4.2=-32\)
gricha - v6959 - 16.1.13

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