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Mathematics 13 Online
OpenStudy (anonymous):

lim as x approaches +infinity (x^7)+(x^5)+(x^3) all divided by (.5^x)

OpenStudy (anonymous):

Try using l'Hopital's rule until the numerator breaks. I haven't tried it directly yet, but that would be my first instinct.

OpenStudy (anonymous):

Thanks but I'm pretty sure that L'Hopital's Rule isnt being used in the packet of problems I was assigned. Any other suggestions?

OpenStudy (anonymous):

original equation = (x^7 + x^5 + x^3)*(2^x) which approaches infty as x --> infty

OpenStudy (anonymous):

@awandererscry how did u solve it? and it was .5^x as the denominator. unless ur saying .5^x as a denominator = 2^x as a numerator??

OpenStudy (anonymous):

0.5^x = (1/2)^x = 1/(2^x) so if you divide by .5^x, you are multiplying the reciprocal which is 2^x

OpenStudy (anonymous):

ok. i think im getting it now. thanks! :)

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