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lfdahl
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Find
\[\lim_{n\rightarrow \infty}\int_{0}^{\infty}\frac{e^{-x}\cos x}{\frac{1}{n}+nx^2}dx.\]
\[\lim_{n\rightarrow \infty}\int_{0}^{\infty}\frac{e^{-x}\cos x}{\frac{1}{n}+nx^2}dx.\]
Thankyou, Klaas van Aarsen, for your participation and a correct answer! (Yes)An alternative solution:Klaas van Aarsen said:Nice one.
My attempt:
Partial integration gives us:
\begin{aligned}\int\frac{e^{-x}\cos x}{\frac{1}{n}+nx^2}\,dx
&= \int e^{-x}\cos x\cdot \frac{n}{1+(nx)^2}\,dx \\
&= \int e^{-x}\cos x\,d\big(\arctan(nx)\big) \\
&= e^{-x}\cos x\arctan(nx) - \int\arctan(nx)\,d\big(e^{-x}\cos x\big)
\end{aligned}
Therefore:
\begin{aligned}\lim_{n\rightarrow \infty}\int_{0}^{\infty}\frac{e^{-x}\cos x}{\frac{1}{n}+nx^2}\,dx
&= \lim_{n\rightarrow \infty}\lim_{a\to 0^+,b\to\infty}\left[e^{-x}\cos x\arctan(nx)\Big|_a^b - \int_a^b\arctan(nx)\,d\big(e^{-x}\cos x\big)\right] \\
&= \lim_{n\rightarrow \infty}\lim_{a\to 0^+,b\to\infty}\left[ - \int_a^b\arctan(nx)\,d\big(e^{-x}\cos x\big)\right] \\
&= \lim_{a\to 0^+,b\to\infty}\left[ - \int_a^b\lim_{n\rightarrow \infty}\arctan(nx)\,d\big(e^{-x}\cos x\big)\right] \\
&= \lim_{a\to 0^+,b\to\infty}\left[ - \int_a^b\frac\pi 2\,d\big(e^{-x}\cos x\big)\right] \\
&= \lim_{a\to 0^+,b\to\infty}\left[ - \frac\pi 2\,e^{-x}\cos x\right]_a^b \\
&=\frac \pi 2
\end{aligned}
The limit of the integral challenge is equal to 0.
To solve for the limit, we can use the properties of limits and the fundamental theorem of calculus. We can also use L'Hopital's rule to simplify the expression and then evaluate the limit.
The limit is equal to 0 because as x approaches infinity, the numerator (e^(-x)cosx) approaches 0 while the denominator (1/n + nx^2) approaches infinity. This results in the overall expression approaching 0.
No, substitution cannot be used to evaluate the limit in this case. The expression contains both a variable in the numerator and the denominator, and substitution is only applicable when the variable is in the numerator.
The value of n affects the limit by changing the rate at which the denominator approaches infinity. As n increases, the denominator increases at a faster rate, resulting in the limit approaching 0 faster.