This limit is a bit more difficult?

Question

lim x--> infinite x^4/(e^(x^2+x+1))
I have to determine its value with l'Hospital rules,and I'm not sure what lim x-->infinite of e^x is.

Answer 1

lim[x → ∞] [x^4 / e^(x^2 + x + 1)]

L'Hopital's rule states that if the limit has an indeterminate form (ie 0/0 or ∞/∞), then you can take the derivative of the numerator and denominator separately to solve. Ie.

lim[x → n] f(x) / g(x) = lim[x → n] f'(x) / g'(x)

If and only if f(x) / g(x) is an indeterminate form.

So, using that rule:

lim[x → ∞] [x^4 / e^(x^2 + x + 1)]

lim[x → ∞] [4x^3 / [(2x + 1)e^(x^2 + x + 1)]]

Still indeterminate, so continue to use L'Hopital's Rule until you have a determinate form. Note that since you have a polynomial expression on top, you will always be able to get a determine form.

lim[x → ∞] [12x^2 / [(2x + 1)^2 * e^(x^2 + x + 1)]]

lim[x → ∞] [24x / [(2x + 1)^3 * e^(x^2 + x + 1)]]

lim[x → ∞] [24 / [(2x + 1)^4 * e^(x^2 + x + 1)]]

Now, substituting you get 24 / ∞ which = 0.

So: lim[x → ∞] [x^4 / e^(x^2 + x + 1)] = 0

Hope that helps =)

Answer 2

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Answer 3

If x=>+infinity e^x=>+infinity
ifx=>-infinity e^x=> 0

Answer 4

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Answer 5

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Answer 6

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Answer 7

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