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Calculus 6.08h - Examples 7 - 9
Three example problems involving l'Hopital's Rule.
7. limit as x approaches 2 of (x² - 4) / (x - 2)
8. limit as x approaches π/2 of (cos 2x + 1) / (sin x - 1)
9. limit as x approaches 0 of (3x² - 1) / (x - 1)
Three example problems involving l'Hopital's Rule.
7. limit as x approaches 2 of (x² - 4) / (x - 2)
8. limit as x approaches π/2 of (cos 2x + 1) / (sin x - 1)
9. limit as x approaches 0 of (3x² - 1) / (x - 1)
Пікірлер: 3
The second example doesn't need l'hoptal's, just use cos(2x) = cos²x - sin²x = 1 - sin²x - sin²x = 1 - 2sin²x And you end up with: -2(1-sin²x)/[(1-sin(x)] = -2[1+sin(x)] which evaluated at pi/2 is equal to -2(1+1) = -4
@derekowens
7 жыл бұрын
Nice! And that's right. Simplifying the expression will also work, when it can be simplified. l'Hopital's Rule works as well, but it isn't the only valid approach.
@lloydgush
7 жыл бұрын
way quicked with l'hopital though. Which is frequently how it's used, saving time.