MATEMATIKA OLIMPIADE: Limit Eksponen

$\small \dpi{100} \fn_jvn \lim_{x \to \infty }\left ( \frac{x+c}{x-c}\right )^{x}=9$
maka c= ...

Jawab :
mengingat
$\small \dpi{100} \fn_jvn \lim_{x \to \infty }\left ( 1+\frac{1}{x} \right )^{x}=e$
atau

$\small \dpi{100} \fn_jvn \lim_{x \to \infty }\left ( 1+\frac{1}{x+a} \right )^{x}=e$
maka
$\small \dpi{100} \fn_jvn \lim_{x \to \infty }\left ( 1+\frac{p}{x} \right )^{x}=e^{x}$
atau
$\small \dpi{100} \fn_jvn \lim_{x \to \infty }\left ( 1+\frac{p}{x+a} \right )^{x}=e^{x}$
dengan demikian soal di atas bisa ditulis menjadi
$\small \dpi{100} \fn_jvn \lim_{x \to \infty }\left ( \frac{x-c}{x-c}+\frac{2c}{x-c} \right )^{x}=9$
$\small \dpi{100} \fn_jvn \lim_{x \to \infty }\left ( 1+\frac{2c}{x-c} \right )^{x}=9$
$\small \dpi{100} \fn_jvn e^{2c}=9\Leftrightarrow 2c=ln9\Leftrightarrow c=\frac{1}{2}ln 9= ln 3$

MATEMATIKA OLIMPIADE

Saturday, September 10, 2011

Limit Eksponen

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