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anonymous

  • 5 years ago

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

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  1. anonymous
    • 5 years ago
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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.

  2. anonymous
    • 5 years ago
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    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?

  3. anonymous
    • 5 years ago
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    original equation = (x^7 + x^5 + x^3)*(2^x) which approaches infty as x --> infty

  4. anonymous
    • 5 years ago
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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??

  5. anonymous
    • 5 years ago
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    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

  6. anonymous
    • 5 years ago
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    ok. i think im getting it now. thanks! :)

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