Find the value of the constant C for which the integral converges: integral from 0 to infinity ((2x/(x^2 +5)) - (C/(3x+2)) dx I tried to integrate it, which led me to ln[(x^2+5)/(3x+2)^(C/3) evaluated at 0 and infinity, but I don't know what to do next.
Find the value of this expression:: \[\large\lim_{x\to\infty}\ln\frac{x^2+5}{(3x+2)^{C/3}}-\lim_{x\to0^+}\ln\frac{x^2+5}{(3x+2)^{C/3}}\] which is a direct result of the fundamental theorem of calculus.
Am I supposed to use L'Hopital's rule?
Sure, L'Hopital's could work. Using the fact that the natural log function is continuous for \(x>0\), you have \[\lim_{x\to\infty}\ln f(x)=\ln\left(\lim_{x\to\infty}f(x)\right)\] and so you can apply L'H to \[\lim_{x\to\infty}\frac{x^2+5}{(3x+2)^{C/3}}\]
surely 3/(3x+2) =ln(3x+2) hence C=3 that's my only thought the rest I don't know
After the rule I get 2x/(C/3)(3x+2)^(C/3-1)
Without using L'Hopital's, you can make use of the fact that the limit of a ratio of polynomials as \(x\to\infty\) is a matter of checking the degrees of the polynomials in the numerator and denominator. If the rational function is \(R(x)=\dfrac{P(x)}{Q(x)}\), where \(P(x)=a_nx^n+\cdots+a_1x+a_0\) and \(Q(x)=b_mx^m+\cdots+b_1x+b_0\), then \[\large\lim_{x\to\infty}R(x)=\begin{cases}\infty&\text{if }n>m\\\\ \dfrac{a_n}{b_m}&\text{if }n=m\\\\ 0&\text{if }n<m\end{cases}\]
Okay then I want to make C/3 equal to 2?
And so C is 6?
Right, \(C=6\).
Okay. Thank you!
yw
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