De som van de limieten

\(\small\displaystyle\lim_{x\to1}\frac{\sqrt x-1}{x-1}\;\;en\;\;\displaystyle\lim_{x\to1}\frac{x^2-1}{x-1}\)

is gelijk aan
A.   1
B.   2,5
C.   0
D.   4
E.   geen van de vorige
A    B    C    D    E

[ 5-6969 - op net sinds 23.3.13-(E)-19.10.2026 ]

Translation in   E N G L I S H

The sum of the limits

\(\small\displaystyle\lim_{x\to1}\frac{\sqrt x-1}{x-1}\;\;and\;\;\displaystyle\lim_{x\to1}\frac{x^2-1}{x-1}\)

is equal to
A.  1
B.  2.5
C.  0
D.  4
E.  none of the above

Oplossing - Solution

1ste manier :
\(\displaystyle\lim_{x\to1}\frac{\sqrt x-1}{x-1}\;+\;\displaystyle \lim_{x\to 1}\frac{x^2-1}{x-1}=\displaystyle\lim_{x \to 1}\frac{(\sqrt x-1)(\sqrt x+1)}{(x-1)(\sqrt x+1)}+\displaystyle \lim_{x \to 1}\frac{(x^2-1)(x+1)}{(x-1)(x+1)}\\=\displaystyle \lim_{x \to 1}\frac{1}{\sqrt x+1}+\displaystyle \lim_{x \to 1}\frac{x+1}{1}=\frac12+2=\frac52\)
2de manier :
\(\displaystyle \lim_{x \to 1}\frac{\sqrt x-1}{x-1}\;+\;\displaystyle \lim_{x \to 1}\frac{x^2-1}{x-1}\overset{H}{=}\displaystyle \lim_{x\to1}\frac{\frac1{2\sqrt x}}{1}+\displaystyle \lim_{x \to 1}\frac{2x}{1}=\frac12+2=\frac52\)
3de manier :
\(\displaystyle\lim_{x\to1}\frac{\sqrt x-1}{x-1}\;+\;\displaystyle\lim_{x\to1}\frac{x^2-1}{x-1}=\displaystyle\lim_{x\to1}\frac{\sqrt x+x^2-2}{x-1}=\displaystyle \lim_{x\to 1}\frac{x^2-1+\sqrt x-1}{x-1}\\=\displaystyle \lim_{x\to 1}\frac{(x-1)(x+1)+(\sqrt x-1)}{x-1}=\displaystyle\lim_{x\to 1}\frac{(\sqrt x-1)(\sqrt x+1)(x+1)+(\sqrt x-1)}{(\sqrt x-1)(\sqrt x+1)}\\=\displaystyle\lim_{x\to1}\frac{(\sqrt x+1)(x+1)+1}{\sqrt x+1}=\frac{2.2+1}{2}=\frac52\)
gricha - v6969 - 23.3.13
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