Matematica - Siruri de numere reale. Limite de functii

Siruri de numere reale. Limite de functii - Intrebarea 11

Mergi la:

Să se calculeze $limit as x rightwards arrow 0 of open parentheses 1 minus sin squared x close parentheses to the power of 5 over x squared end exponent$ .

e to the power of negative 5 end exponent

e to the power of 5

e to the power of negative 1 divided by 5 end exponent

Raspunsul este corect.

Raspunsul este gresit.

Explicatie:

Răspuns corect: A

Nedeterminare $1 to the power of infinity$ :

$limit as x rightwards arrow 0 of open parentheses 1 minus sin squared x close parentheses to the power of 5 over x squared end exponent equals limit as x rightwards arrow 0 of open parentheses 1 plus open parentheses negative sin squared x close parentheses close parentheses to the power of 5 over x squared end exponent equals e to the power of negative 5 limit as x rightwards arrow 0 of fraction numerator sin squared x over denominator x squared end fraction end exponent equals e to the power of negative 5 end exponent.$