Mathc initiation/Fichiers c : c30ch
Apparence
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c2h.c |
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/* --------------------------------- */
/* save as c2h.c */
/* --------------------------------- */
#include "x_hfile.h"
#include "fh.h"
/* --------------------------------- */
int main(void)
{
double i;
clrscrn();
printf(" Does lim x->0 %s exist ?\n\n", feq);
printf(" Substituing 0 for x gives 0/0.\n");
stop();
clrscrn();
printf(" f : x-> %s\n\n", feq);
printf(" Approximate f(x) by the right,\n");
printf(" for x near 0.\n\n");
for(i=1; i>0.1; i+=-.1)
printf(" f(%+.1f) = %5.3f || f(%+.2f) = %5.6f || f(%+.3f) = %5.8f\n",
i, f(i),
i*.1, f(i*.1),
i*.01,f(i*.01)
);
stop();
clrscrn();
printf(" f : x-> %s\n\n", feq);
printf(" Approximate f(x) by the left,\n");
printf(" for x near 0.\n\n");
for(i=-1; i<-0.1; i+=.1)
printf(" f(%+.1f) = %5.3f || f(%+.2f) = %5.6f || f(%+.3f) = %5.8f\n",
i, f(i),
i*.1, f(i*.1),
i*.01,f(i*.01)
);
stop();
clrscrn();
printf(" With the table we arrive at the following conjecture.\n\n");
printf(" lim x->0 %s = 1\n\n", feq);
stop();
return 0;
}
/* --------------------------------- */
/* --------------------------------- */
On peut obtenir le même résultat en utilisant la Règle de L'Hôpital. [wikipedia].
1 --- (log(1+x))' 1+x 1 1 ----------- = --- = --- et lim x->0 --- = 1 x' 1 1+x 1+x
Remarque :
A ----- (log(1+A*x))' 1+A*x A 1 A ------------- = ----- et lim x->0 ----- --- = --- (B*x)' B 1+A*x B B
Exemple de sortie écran :
Does lim x->0 log(1+x)/x exist ?
Substituing 0 for x gives 0/0.
Press return to continue.
****************************
f : x-> log(1+x)/x
Approximate f(x) by the right,
for x near 0.
f(+1.0) = 0.693 || f(+0.10) = 0.953102 || f(+0.010) = 0.99503309
f(+0.9) = 0.713 || f(+0.09) = 0.957530 || f(+0.009) = 0.99552682
f(+0.8) = 0.735 || f(+0.08) = 0.962013 || f(+0.008) = 0.99602121
f(+0.7) = 0.758 || f(+0.07) = 0.966552 || f(+0.007) = 0.99651625
f(+0.6) = 0.783 || f(+0.06) = 0.971148 || f(+0.006) = 0.99701195
f(+0.5) = 0.811 || f(+0.05) = 0.975803 || f(+0.005) = 0.99750830
f(+0.4) = 0.841 || f(+0.04) = 0.980518 || f(+0.004) = 0.99800532
f(+0.3) = 0.875 || f(+0.03) = 0.985293 || f(+0.003) = 0.99850299
f(+0.2) = 0.912 || f(+0.02) = 0.990131 || f(+0.002) = 0.99900133
f(+0.1) = 0.953 || f(+0.01) = 0.995033 || f(+0.001) = 0.99950033
Press return to continue.
****************************
f : x-> log(1+x)/x
Approximate f(x) by the left,
for x near 0.
f(-1.0) = inf || f(-0.10) = 1.053605 || f(-0.010) = 1.00503359
f(-0.9) = 2.558 || f(-0.09) = 1.047896 || f(-0.009) = 1.00452718
f(-0.8) = 2.012 || f(-0.08) = 1.042270 || f(-0.008) = 1.00402146
f(-0.7) = 1.720 || f(-0.07) = 1.036724 || f(-0.007) = 1.00351642
f(-0.6) = 1.527 || f(-0.06) = 1.031257 || f(-0.006) = 1.00301205
f(-0.5) = 1.386 || f(-0.05) = 1.025866 || f(-0.005) = 1.00250836
f(-0.4) = 1.277 || f(-0.04) = 1.020550 || f(-0.004) = 1.00200535
f(-0.3) = 1.189 || f(-0.03) = 1.015307 || f(-0.003) = 1.00150301
f(-0.2) = 1.116 || f(-0.02) = 1.010135 || f(-0.002) = 1.00100134
f(-0.1) = 1.054 || f(-0.01) = 1.005034 || f(-0.001) = 1.00050033
Press return to continue.
****************************
With the table we arrive at the following conjecture.
lim x->0 log(1+x)/x = 1
Press return to continue.