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[ 5-7097 - op net sinds 28.8.12-(E)-11.2.2024 ]
Translation in E N G L I S H
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Oplossing - Solution
1ste manier : met de kennis van de regel van de l'Hospital en
de formule A³ − B³ = (A − B)(A² + AB + B² )
\buildrel&space;H\over=\lim_{x\to&space;0}\,\frac{\cos&space;x-\frac{1}{\cos^2x}}{\cos&space;x-1}=\lim_{x\to&space;0}\:\frac{\cos^3-1}{\cos^2x(\cos&space;x-1)}\\=\lim_{x\to&space;0}\frac{(\cos&space;x-1)(\cos^2x+\cos&space;x+\!1)}{\cos^2x(\cos&space;x-1)}=\lim_{x\to&space;0}\frac{\cos^2x+\cos&space;x+\!1}{\cos^2x}=\frac{1\!+\!1\!+\!1}{1^2}=3&space; "\\\lim_{x\to 0}\:\frac{\sin x\:-\:\tan x}{\sin x\:-\:x}\left(=\frac00\right)\buildrel H\over=\lim_{x\to 0}\,\frac{\cos x-\frac{1}{\cos^2x}}{\cos x-1}=\lim_{x\to 0}\:\frac{\cos^3-1}{\cos^2x(\cos x-1)}\\=\lim_{x\to 0}\frac{(\cos x-1)(\cos^2x+\cos x+\!1)}{\cos^2x(\cos x-1)}=\lim_{x\to 0}\frac{\cos^2x+\cos x+\!1}{\cos^2x}=\frac{1\!+\!1\!+\!1}{1^2}=3")
2de manier : ook met de kennis van de regel van de l'Hospital
\buildrel&space;H\over=1+\lim_{x\to&space;0}\:\frac{1-\frac{1}{\cos^2x}}{\cos&space;x-1}=1+\lim_{x\to&space;0}\:\frac{\cos^2x-1}{\cos^2x(\cos&space;x-1)}\\=1+\lim_{x\to&space;0}\:\frac{\cos&space;x+1}{\cos^2x}=1+\frac{1+1}{1^2}=1+2=3 "\\\lim_{x\to 0}\:\frac{\sin x\:-\:\tan x}{\sin x\:-\:x}=\lim_{x\to 0}\:\frac{1-\frac{1}{\cos x}}{1-\frac{x}{\sin x}}=\lim_{x\to 0}\:\frac{\cos x-1}{\cos x-x\cot x}\\=\lim_{x\to 0}\:\frac{\cos x-x\cot x+x\cot x-1}{\cos x-x\cot x}=1+\lim_{x\to 0}\:\frac{x\cot x-1}{\cos x-x\cot x}\\=1+\lim_{x\to 0}\:\frac{x\cos x-\sin x}{\sin x\cos x-x\cos x}=1+\lim_{x\to 0}\:\frac{x-\tan x}{\sin x-x}\\=1+\left(=\frac00\right)\buildrel H\over=1+\lim_{x\to 0}\:\frac{1-\frac{1}{\cos^2x}}{\cos x-1}=1+\lim_{x\to 0}\:\frac{\cos^2x-1}{\cos^2x(\cos x-1)}\\=1+\lim_{x\to 0}\:\frac{\cos x+1}{\cos^2x}=1+\frac{1+1}{1^2}=1+2=3")
