\(\displaystyle \lim_{\underset{>}{x \to a}} \frac {\tan x\,-\,\tan x} {x\:-\:a}\)
is gelijk aan
|
A. 1 |
|---|
| B. sec2 a |
| C. csc2 a |
| D. tan2 a |
E.  |
[ 5-6202 - op net sinds 1.2.13-(E)-19.10.2026 ]
Translation in E N G L I S H
|
\(\displaystyle \lim_{x \to a^+} \frac {\tan x\,-\,\tan x} {x\:-\:a}=\)
|
A. 1 |
|---|
| B. sec2a |
| C. csc2a |
| D. tan2a |
| E. ∞ |
Oplossing - Solution
Het maakt niet uit of
\(\underset{>}{ x \to a}\)
of \(\underset{<}{ x \to a}\)
of \({x\to a}\)
1ste manier : zonder de regel van de l'Hospital
\(\displaystyle\lim_{x\to a}\frac{\tan x\,-\,\tan x}{x\:-\:a}=\displaystyle\lim_{x\to a}\frac{\frac{\sin x}{\cos x}-\frac{\sin a}{\cos a}}{x-a}=\displaystyle\lim_{x\to a}\frac{\frac{\sin x.\cos a-\sin a.\cos x}{\cos x.\cos a}}{x-a}=\displaystyle\lim_{x\to a}\frac{\frac{\sin(x-a)}{\cos x.\cos a}}{x-a}\\=\displaystyle\lim_{x\to a}\frac{\sin(x-a)}{x-a}.\displaystyle\lim_{x\to a}\frac{1}{\cos x.\cos a}=1.\frac{1}{\cos a.\cos a}=\frac{1}{\cos^2a}=sec^2a\)
2de manier : met de regel van de L'Hospital
\(\displaystyle\lim_{x\to a}\frac{\tan x-\tan a}{x-a}\left(=\frac00\right)\overset{H}{=}\displaystyle\lim_{x\to a}\frac{\frac{1}{\cos^2x}}{1}=\frac{1}{\cos^2a}=\sec^2a\)
