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

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

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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Try using l'Hopital's rule until the numerator breaks. I haven't tried it directly yet, but that would be my first instinct.
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?
original equation = (x^7 + x^5 + x^3)*(2^x) which approaches infty as x --> infty

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

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