MHB Evaluate Limit: $$\lim_{x\to\infty} (-1)^nn^3 + 2^{-n}$$

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The limit $$\lim_{n\to\infty} (-1)^n n^3 + 2^{-n}$$ is being evaluated, with the conclusion that L'Hopital's rule is not applicable. The term $2^{-n}$ approaches 0 as n approaches infinity. However, the term $(-1)^n n^3$ represents an alternating sequence of cubes, which diverges. Therefore, the overall limit does not exist due to the divergence of the first term. The final result is that the limit is undefined.
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I have this limit:

$$\lim_{{x}\to{\infty}} {(-1)}^{n}{n}^{3} + {2}^{-n}$$

and I'm unsure how to evaluate it or how to apply L'hopital's rule to this limit.
 
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tmt said:
I have this limit: $$\lim_{n\to\infty} (-1)^n\,n^3 + 2^{-n}$$

and I'm unsure how to evaluate it or how to apply L'hopital's rule to this limit.
L'Hopital's rule doesn't apply here.

We see that: \lim_{n\to\infty}2^{-n}\:=\:0

But the first part is an alternating sequence of cubes:
. . -1 + 8 - 27 + 64 - 125 + \cdots which diverges.

 
It's a limit, not a series - the limit does not exist.
 
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