MHB Area under curves and Limit of a sequence,

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Francesco seeks assistance with calculating the area under a curve using the integral from 0 to 2 of the expression \(-(x-2)^3 + 2 - (x^2 - 2)\). The discussion highlights that limits of the form \(\lim_{n\to\infty}f(n)^{g(n)}\) can be simplified using the expression \(e^{g(n)\ln(f(n))}\). Various methods for finding limits are mentioned, including substitution, expansion, and l'Hospital's rule. It is noted that l'Hospital's rule requires the function to be expressed as a ratio of two functions that approach zero or infinity. The conversation emphasizes the importance of understanding these techniques for solving the problems efficiently.
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Hello, I am looking for an help about this, I have very short time to do many of them and those are an example, could someone show me one solution or explain me how to do it?
Thank you if you can help me, I really appreciate.
Francesco.

 
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Judging by the picture (clickable)

[GRAPH]5wib4rocqz[/GRAPH]

the area is
\[
\int_{0}^2(-(x-2)^3+2)-(x^2-2)\,dx
\]

Limits of the form $\lim_{n\to\infty}f(n)^{g(n)}$ are usually easier to compute when the function is represented as $e^{g(n)\ln(f(n))}$. Which ways of finding limits do you know?
 
Evgeny.Makarov said:
Judging by the picture (clickable)

[GRAPH]5wib4rocqz[/GRAPH]

the area is
\[
\int_{0}^2(-(x-2)^3+2)-(x^2-2)\,dx
\]

Limits of the form $\lim_{n\to\infty}f(n)^{g(n)}$ are usually easier to compute when the function is represented as $e^{g(n)\ln(f(n))}$. Which ways of finding limits do you know?

By substitution, or expanding, or Hospital rule i suppose.

Thank you
 
namerequired said:
By substitution, or expanding, or Hospital rule i suppose.
I think, the easiest way is to expand $\ln(1+x)$ as $x+o(x)$, but l'Hospital's rule works too. Recall that to apply the rule you need to represent the function as a ratio of two functions that tend both to zero or both to infinity.
 

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