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$\textbf{Compute the bounderies:} \int_1^{\infty} \frac{6x}{x^4 + 6x^2 + 4} = 0 - \big(-\ln\big(\frac{5}{2}\big)\big)$

$\int_a^b f(x)dx = F(b) - F(a) = \lim_{x\to b^-}(F(x)) - \lim_{x\to a^+} (F(x))$

$\lim_{x\to 1^+} \bigg(-\ln\bigg|\frac{8}{3} + \frac{2x^2}{3}\bigg| + \ln\bigg|\frac{2}{3} + \frac{2x^2}{3}\bigg|\bigg) = -\ln(\frac{5}{2})$

$\lim_{x\to\infty} \bigg(-\ln\bigg|\frac{8}{3} + \frac{2x^2}{3}\bigg| + \ln\bigg|\frac{2}{3} + \frac{2x^2}{3}\bigg|\bigg) = 0$

$ = 0 - \big(-\ln\big(\frac{5}{2}\big)\big)$

Can someone provide an explanation for why the second limit is equal to 0?

Only thing I can evaluate it to at the moment is -$\infty$ + $\infty$ which is not 0

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    $\begingroup$ Nowhere in that picture is there any assertion to the effect that $-\infty+\infty=0$. $\endgroup$ Commented Nov 5, 2017 at 17:33
  • $\begingroup$ Can you elaborate on how to evaluate that limit? Because the only thing I can evaluate it to is -inf + inf which I know is not 0 $\endgroup$
    – Smit Shah
    Commented Nov 5, 2017 at 17:35
  • $\begingroup$ Is it just an argument that the 8/3 and 2/3 are insignificant and therefore -ln(2x^2 /3) + ln(2x^2 /3) are equal to 0? $\endgroup$
    – Smit Shah
    Commented Nov 5, 2017 at 17:36

1 Answer 1

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We can get rid of the absolute values since $x$ is headed to positive infinity. So,

$$-\ln\bigg(\frac{2}{3} + \frac{2x}{3}\bigg) + \ln\bigg(\frac{8}{3} + \frac{2x^2}{3}\bigg) = \ln\Bigg({\frac{\frac{2}{3} + \frac{2x^2}{3}}{\frac{8}{3} + \frac{2x^2}{3}}}\Bigg)= \ln\bigg(1 - \frac{3}{x^2 + 4}\bigg)$$

$$\lim_{x\to \infty} \ln\bigg(1 - \frac{3}{x^2 + 4}\bigg) = \ln(1) = 0$$

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  • $\begingroup$ Can you explain how you got to (1 - 3/(x^2 + 4)) $\endgroup$
    – Smit Shah
    Commented Nov 5, 2017 at 17:45
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    $\begingroup$ $\frac{\frac{2x^2 + 2}{3}}{\frac{2x^2 + 8}{3}} = \frac{2x^2 + 2}{2x^2 + 8} = \frac{x^2 + 1}{x^2 + 4}$ Then you use polynomial division and see that $x^2 + 4$ divides $x^2 + 1$ once, with a remainder of $\frac{-3}{x^2 + 4}$. Therefore, $\frac{\frac{2x^2 + 2}{3}}{\frac{2x^2 + 8}{3}} = 1-\frac{3}{x^2 + 4}$ $\endgroup$
    – G_D
    Commented Nov 5, 2017 at 17:58

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