13.4B Limieten bij logaritmische functies

Ik kan de limiet van een logaritmische functie bepalen.

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This lesson contains 28 slides, with text slides.

Lesson duration is: 45 min

Items in this lesson

13.4B Limieten bij logaritmische functies

Ik kan de limiet van een logaritmische functie bepalen.

Slide 1 - Slide

Standaardlimieten

Slide 2 - Slide

Hier zit een probleem

Slide 3 - Slide

Geef de volgende elementen:

geogebra

f (x) = \frac{​ 5 + ln ( x ) ​ ​}{​ 2 ln ( x ) - 3 ​}

​ x ↓ 0 ​ lim ​ ​ f (x)

​ x \to \infty ​ lim ​ ​ f (x)

V.A.

Slide 4 - Slide

f (x) = \frac{​ 5 + ln ( x ) ​ ​}{​ 2 ln ( x ) - 3 ​}

​ x ↓ 0 ​ lim ​ ​ f (x)

Slide 5 - Slide

Dus

f (x) = \frac{​ 5 + ln ( x ) ​ ​}{​ 2 ln ( x ) - 3 ​}

​ x ↓ 0 ​ lim ​ ​ f (x)

​ x ↓ 0 ​ lim ​ ​ \frac{​ 5 + ln ( x ) ​ ​}{​ 2 ln ( x ) - 3 ​} = ​ ln (x) \to - \infty ​ lim ​ ​ \frac{​ 5 + ln ( x ) ​ ​}{​ 2 ln ( x ) - 3 ​}

Slide 6 - Slide

f (x) = \frac{​ 5 + ln ( x ) ​ ​}{​ 2 ln ( x ) - 3 ​}

​ x ↓ 0 ​ lim ​ ​ f (x)

​ x ↓ 0 ​ lim ​ ​ \frac{​ 5 + ln ( x ) ​ ​}{​ 2 ln ( x ) - 3 ​} = ​ ln (x) \to - \infty ​ lim ​ ​ \frac{​ 5 + ln ( x ) ​ ​}{​ 2 ln ( x ) - 3 ​}

​ ln (x) \to - \infty ​ lim ​ ​ \frac{​ \frac{​ 5 ​ ​}{​ ln ( x ) ​} + 1 ​ ​}{​ 2 - \frac{​ 3 ​ ​}{​ ln ( x ) ​} ​}

Slide 7 - Slide

f (x) = \frac{​ 5 + ln ( x ) ​ ​}{​ 2 ln ( x ) - 3 ​}

​ x ↓ 0 ​ lim ​ ​ f (x)

​ x ↓ 0 ​ lim ​ ​ \frac{​ 5 + ln ( x ) ​ ​}{​ 2 ln ( x ) - 3 ​} = ​ ln (x) \to - \infty ​ lim ​ ​ \frac{​ 5 + ln ( x ) ​ ​}{​ 2 ln ( x ) - 3 ​}

​ ln (x) \to - \infty ​ lim ​ ​ \frac{​ \frac{​ 5 ​ ​}{​ ln ( x ) ​} + 1 ​ ​}{​ 2 - \frac{​ 3 ​ ​}{​ ln ( x ) ​} ​} = \frac{​ 0 + 1 ​ ​}{​ 2 - 0 ​} = \frac{​ 1 ​ ​}{​ 2 ​}

Slide 8 - Slide

f (x) = \frac{​ 5 + ln ( x ) ​ ​}{​ 2 ln ( x ) - 3 ​}

​ x \to \infty ​ lim ​ ​ f (x)

Slide 9 - Slide

Dus

f (x) = \frac{​ 5 + ln ( x ) ​ ​}{​ 2 ln ( x ) - 3 ​}

​ x \to \infty ​ lim ​ ​ \frac{​ 5 + ln ( x ) ​ ​}{​ 2 ln ( x ) - 3 ​} = ​ ln (x) \to \infty ​ lim ​ ​ \frac{​ 5 + ln ( x ) ​ ​}{​ 2 ln ( x ) - 3 ​}

​ x \to \infty ​ lim ​ ​ f (x)

Slide 10 - Slide

f (x) = \frac{​ 5 + ln ( x ) ​ ​}{​ 2 ln ( x ) - 3 ​}

​ x \to \infty ​ lim ​ ​ \frac{​ 5 + ln ( x ) ​ ​}{​ 2 ln ( x ) - 3 ​} = ​ ln (x) \to \infty ​ lim ​ ​ \frac{​ 5 + ln ( x ) ​ ​}{​ 2 ln ( x ) - 3 ​}

​ ln (x) \to \infty ​ lim ​ ​ \frac{​ \frac{​ 5 ​ ​}{​ ln ( x ) ​} + 1 ​ ​}{​ 2 - \frac{​ 3 ​ ​}{​ ln ( x ) ​} ​}

​ x \to \infty ​ lim ​ ​ f (x)

Slide 11 - Slide

f (x) = \frac{​ 5 + ln ( x ) ​ ​}{​ 2 ln ( x ) - 3 ​}

​ x \to \infty ​ lim ​ ​ \frac{​ 5 + ln ( x ) ​ ​}{​ 2 ln ( x ) - 3 ​} = ​ ln (x) \to \infty ​ lim ​ ​ \frac{​ 5 + ln ( x ) ​ ​}{​ 2 ln ( x ) - 3 ​}

​ ln (x) \to \infty ​ lim ​ ​ \frac{​ \frac{​ 5 ​ ​}{​ ln ( x ) ​} + 1 ​ ​}{​ 2 - \frac{​ 3 ​ ​}{​ ln ( x ) ​} ​} = \frac{​ 0 + 1 ​ ​}{​ 2 - 0 ​} = \frac{​ 1 ​ ​}{​ 2 ​}

​ x \to \infty ​ lim ​ ​ f (x)

Slide 12 - Slide

V. A.

f (x) = \frac{​ 5 + ln ( x ) ​ ​}{​ 2 ln ( x ) - 3 ​}

Slide 13 - Slide

V. A.

f (x) = \frac{​ 5 + ln ( x ) ​ ​}{​ 2 ln ( x ) - 3 ​}

2 ln (x) - 3 = 0

Slide 14 - Slide

V. A.

f (x) = \frac{​ 5 + ln ( x ) ​ ​}{​ 2 ln ( x ) - 3 ​}

2 ln (x) - 3 = 0

2 ln (x) = 3

Slide 15 - Slide

V. A.

f (x) = \frac{​ 5 + ln ( x ) ​ ​}{​ 2 ln ( x ) - 3 ​}

2 ln (x) - 3 = 0

2 ln (x) = 3

ln (x) = 1 \frac{​ 1 ​ ​}{​ 2 ​}

Slide 16 - Slide

V. A.

f (x) = \frac{​ 5 + ln ( x ) ​ ​}{​ 2 ln ( x ) - 3 ​}

2 ln (x) - 3 = 0

2 ln (x) = 3

ln (x) = 1 \frac{​ 1 ​ ​}{​ 2 ​}

x = e^{​ 1 \frac{​ 1 ​ ​}{​ 2 ​} ​ ​}

Slide 17 - Slide

V. A.

f (x) = \frac{​ 5 + ln ( x ) ​ ​}{​ 2 ln ( x ) - 3 ​}

2 ln (x) - 3 = 0

2 ln (x) = 3

ln (x) = 1 \frac{​ 1 ​ ​}{​ 2 ​}

x = e^{​ 1 \frac{​ 1 ​ ​}{​ 2 ​} ​ ​} = e \sqrt ​ e ​ ​ ​

Slide 18 - Slide

Geef de volgende elementen:

​ x ↓ 0 ​ lim ​ ​ f (x)

​ x \to \infty ​ lim ​ ​ f (x)

V.A.

g (x) = \frac{​ 5 \cdot ^{​ 0, 5 ​ ​} lo g ( x ) - 4 ​ ​}{​ 4 - ^{​ 0, 5 ​ ​} lo g ( x ^{​ 2 ​ ​} ) ​}

Slide 19 - Slide

V. A.

geogebra

g (x) = \frac{​ 5 \cdot ^{​ 0, 5 ​ ​} lo g ( x ) - 4 ​ ​}{​ 4 - ^{​ 0, 5 ​ ​} lo g ( x ^{​ 2 ​ ​} ) ​}

Slide 20 - Slide

V. A.

geogebra

x = \frac{​ 1 ​ ​}{​ 4 ​}

g (x) = \frac{​ 5 \cdot ^{​ 0, 5 ​ ​} lo g ( x ) - 4 ​ ​}{​ 4 - ^{​ 0, 5 ​ ​} lo g ( x ^{​ 2 ​ ​} ) ​}

Slide 21 - Slide

​ x ↓ 0 ​ lim ​ ​ \frac{​ 5 \cdot ^{​ 0, 5 ​ ​} lo g ( x ) - 4 ​ ​}{​ 4 - ^{​ 0, 5 ​ ​} lo g ( x ^{​ 2 ​ ​} ) ​}

Slide 22 - Slide

​ x ↓ 0 ​ lim ​ ​ \frac{​ 5 \cdot ^{​ 0, 5 ​ ​} lo g ( x ) - 4 ​ ​}{​ 4 - ^{​ 0, 5 ​ ​} lo g ( x ^{​ 2 ​ ​} ) ​} = ​^{​ 0, 5 ​ ​} lo g (x) \to \infty ​ lim ​ ​ \frac{​ 5 \cdot ^{​ 0, 5 ​ ​} lo g ( x ) - 4 ​ ​}{​ 4 - ^{​ 0, 5 ​ ​} lo g ( x ^{​ 2 ​ ​} ) ​}

Slide 23 - Slide

​ x ↓ 0 ​ lim ​ ​ \frac{​ 5 \cdot ^{​ 0, 5 ​ ​} lo g ( x ) - 4 ​ ​}{​ 4 - ^{​ 0, 5 ​ ​} lo g ( x ^{​ 2 ​ ​} ) ​} = ​^{​ 0, 5 ​ ​} lo g (x) \to \infty ​ lim ​ ​ \frac{​ 5 \cdot ^{​ 0, 5 ​ ​} lo g ( x ) - 4 ​ ​}{​ 4 - ^{​ 0, 5 ​ ​} lo g ( x ^{​ 2 ​ ​} ) ​}

​^{​ 0, 5 ​ ​} lo g (x) \to \infty ​ lim ​ ​ \frac{​ 5 \cdot ^{​ 0, 5 ​ ​} lo g ( x ) - 4 ​ ​}{​ 4 - 2 \cdot ^{​ 0, 5 ​ ​} lo g ( x ) ​}

Slide 24 - Slide

​^{​ 0, 5 ​ ​} lo g (x) \to \infty ​ lim ​ ​ \frac{​ 5 \cdot ^{​ 0, 5 ​ ​} lo g ( x ) - 4 ​ ​}{​ 4 - 2 \cdot ^{​ 0, 5 ​ ​} lo g ( x ) ​} = ​^{​ 0, 5 ​ ​} lo g (x) \to \infty ​ lim ​ ​ \frac{​ 5 - \frac{​ 4 ​ ​}{​ ^{​ 0, 5 ​ ​} lo g ( x ) ​} ​ ​}{​ \frac{​ 4 ​ ​}{​ ^{​ 0, 5 ​ ​} lo g ( x ) ​} - 2 ​}

​ x ↓ 0 ​ lim ​ ​ \frac{​ 5 \cdot ^{​ 0, 5 ​ ​} lo g ( x ) - 4 ​ ​}{​ 4 - ^{​ 0, 5 ​ ​} lo g ( x ^{​ 2 ​ ​} ) ​} = ​^{​ 0, 5 ​ ​} lo g (x) \to \infty ​ lim ​ ​ \frac{​ 5 \cdot ^{​ 0, 5 ​ ​} lo g ( x ) - 4 ​ ​}{​ 4 - ^{​ 0, 5 ​ ​} lo g ( x ^{​ 2 ​ ​} ) ​}

Slide 25 - Slide

​^{​ 0, 5 ​ ​} lo g (x) \to \infty ​ lim ​ ​ \frac{​ 5 \cdot ^{​ 0, 5 ​ ​} lo g ( x ) - 4 ​ ​}{​ 4 - 2 \cdot ^{​ 0, 5 ​ ​} lo g ( x ) ​} = ​^{​ 0, 5 ​ ​} lo g (x) \to \infty ​ lim ​ ​ \frac{​ 5 - \frac{​ 4 ​ ​}{​ ^{​ 0, 5 ​ ​} lo g ( x ) ​} ​ ​}{​ \frac{​ 4 ​ ​}{​ ^{​ 0, 5 ​ ​} lo g ( x ) ​} - 2 ​} =

\frac{​ 5 - 0 ​ ​}{​ 0 - 2 ​} = - 2 \frac{​ 1 ​ ​}{​ 2 ​}

​ x ↓ 0 ​ lim ​ ​ \frac{​ 5 \cdot ^{​ 0, 5 ​ ​} lo g ( x ) - 4 ​ ​}{​ 4 - ^{​ 0, 5 ​ ​} lo g ( x ^{​ 2 ​ ​} ) ​} = ​^{​ 0, 5 ​ ​} lo g (x) \to \infty ​ lim ​ ​ \frac{​ 5 \cdot ^{​ 0, 5 ​ ​} lo g ( x ) - 4 ​ ​}{​ 4 - ^{​ 0, 5 ​ ​} lo g ( x ^{​ 2 ​ ​} ) ​}

Slide 26 - Slide

Toon nu op dezelfde wijze aan

​ x \to \infty ​ lim ​ ​ g (x)

Slide 27 - Slide

​ x \to \infty ​ lim ​ ​ \frac{​ 5 \cdot ^{​ 0, 5 ​ ​} lo g ( x ) - 4 ​ ​}{​ 4 - ^{​ 0, 5 ​ ​} lo g ( x ^{​ 2 ​ ​} ) ​} = ​^{​ 0, 5 ​ ​} lo g (x) \to - \infty ​ lim ​ ​ \frac{​ 5 \cdot ^{​ 0, 5 ​ ​} lo g ( x ) - 4 ​ ​}{​ 4 - ^{​ 0, 5 ​ ​} lo g ( x ^{​ 2 ​ ​} ) ​}

​^{​ 0, 5 ​ ​} lo g (x) \to - \infty ​ lim ​ ​ \frac{​ 5 \cdot ^{​ 0, 5 ​ ​} lo g ( x ) - 4 ​ ​}{​ 4 - 2 \cdot ^{​ 0, 5 ​ ​} lo g ( x ) ​} = ​^{​ 0, 5 ​ ​} lo g (x) \to - \infty ​ lim ​ ​ \frac{​ 5 - \frac{​ 4 ​ ​}{​ ^{​ 0, 5 ​ ​} lo g ( x ) ​} ​ ​}{​ \frac{​ 4 ​ ​}{​ ^{​ 0, 5 ​ ​} lo g ( x ) ​} - 2 ​} =

\frac{​ 5 - 0 ​ ​}{​ 0 - 2 ​} = - 2 \frac{​ 1 ​ ​}{​ 2 ​}

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13.4B Limieten bij logaritmische functies

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