I think if we let $t=\frac1z$, then $$\lim_{z\to\infty}\frac{\sin(z)}z=\lim_{t\to0}t\sin(\frac1t)=0$$ but I don't know why in Ablowitz-Fokas Complex Variables, in "Answer to Odd-Numbered Exercises", answer is "doesn't exist"?
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Hint: Wolfram|Alpha agrees with you. – Mar 14 '14 at 06:52
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@Emracool wolfram alpha assumes that $z$ is real. – mrf Mar 14 '14 at 06:54
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@mrf Good point. Didn't catch that one. – Mar 14 '14 at 06:55
3 Answers
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Note that $z$ and hence your $t$ is not necessarily real. Investigate what happens when you let $z$ approach $\infty$ along the imaginary axis.
mrf
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Do you say that I let $z=iy$ and evaluate $\lim_{y\to\infty}\frac{\sin(iy)}{iy}$? – Mar 14 '14 at 07:02
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Hint: Unlike real sine and cosine, complex sine and cosine are unbounded.
Pick two sequences $a_n$ and $b_n$ such that $\displaystyle \lim_{n\rightarrow\infty}a_n=\lim_{n\rightarrow\infty}b_n=\infty$ but $\displaystyle \lim_{n\rightarrow\infty}\sin(a_n)/a_n$ and $\displaystyle \lim_{n\rightarrow\infty}\sin(b_n)/b_n$ are different.
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Hint: $\sin(ix) = i\sinh(x)$. Therefore if that limit exists then this limit too must exist: $$\displaystyle \lim_{x \to \infty} \frac{\sin(ix)}{ix}=\lim_{x \to \infty} \frac{i\sinh(x)}{ix}$$
math.n00b
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