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Limieten bewijzen

Je moet bewijzen dat 3 de limiet is van de functie:
f(x)=x²-1

lim (x²-1) = 3
x$\to$-2

|(x²-1)-3| $<$ $\epsilon$
|x-(-2)| $<$ $\delta$

Ik kom tot deze stap, maar hoe moet ik nu verder?

Sara
8-9-2012

Antwoord

Ik zal je een voorbeeld geven. Als je dat 's ernstig bestudeert dan kan je 't zelf ook wel, denk ik.

$
\begin{array}{l}
\mathop {\lim }\limits_{x \to 3} x^3 - 2x^2 + 3x - 4 = 14 \\
\left| {f(x) - 14} \right| = \left| {x^3 - 2x^2 + 3x - 4 - 14} \right| = \left| {x^3 - 2x^2 + 3x - 18} \right| = \left| {x - 3} \right| \cdot \left| {...} \right| \\
x - 3/x^3 - 2x^2 + 3x - 18\backslash x^2 + x + 6 \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,x^3 - 3x^2 \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x^2 + 3x - 18 \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x^2 - 3x \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,6x - 18 \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,6x - 18 \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \\
\left| {f(x) - 14} \right| = \left| {x^3 - 2x^2 + 3x - 18} \right| = \left| {x - 3} \right| \cdot \left| {x^2 + x + 6} \right| \\
Kies\,\,\delta _1 = 1\,\,dan\,\,\left| {x - 3} \right| < 1 \\
- 1 < x - 3 < 1,\,\,dus\,\,2 < x < 4,\,\,zodat\,\,\left| x \right| < 4 \\
Met\,\,\left| x \right| < 4\,\,geldt\,\,\left| {x^2 + x + 6} \right| \le 4^2 + 4 + 6 < 26 \\
Kies\,\,\delta _2 = \frac{1}{{26}}\varepsilon \,\,en\,\,neem\,\,\delta = \min (\delta _1 ,\delta _2 ),\,\,dan\,\,geldt: \\
\left| {f(x) - 14} \right| = \left| {x - 3} \right| \cdot \left| {x^2 + x + 6} \right| \le \frac{1}{{26}}\varepsilon \, \cdot 26 = \varepsilon \\
\end{array}
$

Lukt dat zo?

WvR
8-9-2012