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I am wondering how to evaluate the limit in the picture below. I've read online that the limit at infinity is zero when the denominator $\geq$ (numerator $+2$) but why is this? I am skeptical that I can use the Cauchy-Goursat theorem. Is there a way to show that the limit goes to zero using the ML inequality.

Page 169 of the textbook says that if $|f(z)|\leq M$ For all $z$ on a (piece wise smooth) curve $C$ With length $L,$ then $\left|\int_Cf\right|\leq M\cdot L.$
Now let $C$ be the upper half-circle $|z|=R,$ $\operatorname{Im} z>0.$ Show: $$\lim_{R\to\infty}\int_C \frac{2z^2-5}{(z^2+1)(z^2+4)}\,dz=0$$

Thanks

Thomas Andrews
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Elli
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  • Possibly helpful: https://math.stackexchange.com/q/4137201/42969. – Martin R May 30 '21 at 19:13
  • I don't know anything about residues so I'm looking for a different explanation. I think the problem is asking for an ML explanation. – Elli May 30 '21 at 19:15
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    So what do you get when estimating the integral? The length of the curve is $2 \pi R$, and the integrand is $\sim 2/z^2$ for $z \to \infty$. – Martin R May 30 '21 at 19:17
  • Ya that is my thinking as well, but is it that simple. That's pretty much a calculus theorem, and I don't see it relating to the ML theorem. It's 10 points which is also kinda strange – Elli May 30 '21 at 19:19
  • It is that simple. $M \sim 2/R^2$ and $L = 2 \pi R$. – Martin R May 30 '21 at 19:21
  • Thank you for your help Martin. – Elli May 30 '21 at 19:39

2 Answers2

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Hint: Show: $$\begin{align}\left|2z^2-5\right|&\leq 2R^2+5\\ \left|z^2+1\right|&\geq R^2-1\\ \left|z^2+4\right|&\geq R^2-4 \end{align}$$

Thomas Andrews
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This does not require Cauchy-Goursat to estimate. Taking the extreme bounds on the contour of integration, and using that the length of the contour of integration is $\pi R$, we get for $R\gt2$, $$ \begin{align} \left|\,\int_{\!\!\substack{|z|=R\\\mathrm{Im}(z)\ge0}\!\!}\frac{2z^2-5}{(z^2+1)(z^2+4)}\,\mathrm{d}z\,\right| &\le\frac{2R^2+5}{\left(R^2-1\right)\left(R^2-4\right)}\,\pi R \end{align} $$ Finish by taking the limit as $R\to\infty$.

robjohn
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