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how can I proceed ? can I use the same techniques we use when computing a limit where the variable is a real number ?

edit : forgot to mention that z is a complex number

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    $\begingroup$ It is universally known that this is one. $\sin(x)/x$ is the sinc function and a search of that should reveal some proofs. But my favourite way is through the Maclaurin series expansion of the sine - which works for all $x$, real or complex. $\endgroup$ Feb 14, 2017 at 13:55

3 Answers 3

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Hint: $$\lim_{z \rightarrow 0}\frac{\sin(z)-\sin(0)}{z-0}=(\sin)'(0)=\cos(0)=1$$

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    $\begingroup$ add $\lim_{z\to0}$ $\endgroup$
    – Nosrati
    Feb 14, 2017 at 14:06
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One technique is the use of power series:

For $ z \ne 0$: $ \frac{\sin z}{z}=1-\frac{z^2}{3!}+\frac{z^4}{5!}-+... \to 1$ for $z \to 0$

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Maybe a more formal proof:

Denote $f(z)=\frac {\sin(z)}{z}$. $f$ is analytic in $\Bbb C$ except for a singularity point in $0$.

Look at the Taylor series $\sin(z)=\sum_{n=0}^\infty (-1)^n \frac {z^{2n+1}}{(2n+1)!}$. Which implies $\frac{\sin(z)}{z}=\sum_{n=0}^\infty (-1)^n \frac {z^{2n}}{(2n+1)!}$. This is still a Taylor series (there are no addends of the form $\frac{A}{z^{-n}}$ where $A\not= 0, n\in\Bbb N$). This means that $\lim_{z\to z_0}\frac{\sin(z)}{z}=\sum_{n=0}^\infty (-1)^n \frac {z^{2n}}{(2n+1)!}|_{z=0}=1$.

Remark: If you wouldn't have gotten a Taylor series the limit would have been $\infty$ if you would have gotten a series of the structure $\sum_{n=-m}^\infty a_{n}z^n$ and undefined if the series would have been of the structure $\sum_{n=-\infty}^\infty a_{n}z^n$ where $a_{-n}\not=0$ infinitely often for $n\in\Bbb N$.

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  • $\begingroup$ The line with the part "This means that..." must end with a $\;1\;$ ... $\endgroup$
    – DonAntonio
    Feb 14, 2017 at 14:41
  • $\begingroup$ Obviously. Fixed and thanks for noting. $\endgroup$
    – Shaked
    Feb 14, 2017 at 14:49

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