$Convergence integrals$

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$Convergence integrals$

$\int_{15}^{\infty} \frac{-6x^4-7x^2+1}{7x^4-x^3+x}dx $

Calculus Improper Integrals Convergence

Phia

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Note that
\[\lim_{x \rightarrow \infty} \frac{-6x^4-7x^2+1}{7x^4-x^3+x}=-\frac{6}{7}\]
\[=\lim_{x \rightarrow \infty} \frac{x^4(-6-\frac{7}{x^2}+\frac{1}{x^4})}{x^4(7-\frac{1}{x}+\frac{1}{x^3})}\]
\[\lim_{x \rightarrow \infty} \frac{-6-\frac{7}{x^2}+\frac{1}{x^4}}{7-\frac{1}{x}+\frac{1}{x^3}}=\frac{-6}{7}\neq0.\]
Hence the integral is divergent.

Erdos

Erdos

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Leave a comment if you need any clarifications.
Erdos

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@Phia: What details do you need?
Erdos

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I added some more detail. Let me know if you need any clarifications, but the solution is pretty much what I have written.
Phia

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Can you explain why the integral converges if the lim is 0?
Erdos

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I am not saying that he integral converges if the lim is 0. If the limit is not zero then the integral is divergent. This is a well known fact about improper integrals which can be easily found in any calculus textbook.

$Convergence integrals$

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