The derivative of $\sin(\alpha x)/ \sin(x)$ is continuous at $0$? Let $\alpha$ be a real number. Define $f = \sin(\alpha x)/ \sin(x)$ if $x \neq 0$ and $\alpha$ at $x = 0.$ How can I prove the derivative of $\sin(\alpha x)/ \sin(x)$ is continuous at $0$? I took the derivative and tried to apply L'Hopital's rule but I kept getting $0/0$ and could not make the argument work... Any explanation is appreciated.
Both $\operatorname{sinc}(x)$ and its reciprocal are analytic, non-vanishing functions in a neigbourhood of zero, so it is
$$ \alpha\operatorname{sinc}(\alpha x)\frac{1}{\operatorname{sinc}(x)}. $$
An elementary proof without l'Hôpital nor Landau's big O notation:
for all $x\neq 0$,
\begin{align*}
f'(x)&amp;=\frac{\alpha\cos(\alpha x)\sin(x)-\sin(\alpha x)\cos(x)}{\sin^2(x)}\\
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&amp;=\left(\frac{(x-0)}{\sin(x)-\sin(0)}\right)^2\frac{\alpha\cos(\alpha x)\sin(x)-\sin(\alpha x)\cos(x)}{(x-0)^2}.
\end{align*}
By continuity of the square and inverse functions, we have:
\begin{align*}
\lim_{x\to0}\left(\frac{(x-0)}{\sin(x)-\sin(0)}\right)^2&amp;=\left(\frac{1}{\lim_{x\to0}\frac{\sin(x)-\sin(0)}{(x-0)}}\right)^2=\left(\frac{1}{\cos(0)}\right)^2=1.
\end{align*}
Let us write:
\begin{align*}
\alpha\cos(\alpha x)\sin(x)-\sin(\alpha x)\cos(x)&amp;=\alpha\big(\cos(\alpha x)-\cos(0)\big)\sin(x)+\alpha\cos(0)\sin(x)\\
&amp;\quad-\sin(\alpha x)\big(\cos(x)-\cos(0)\big)-\sin(\alpha x)\cos(0)
\end{align*}
Notice first that
\begin{align*}
\lim_{x\to0}\frac{\alpha\big(\cos(\alpha x)-\cos(0)\big)\sin(x)}{(x-0)^2}&amp;=\alpha\lim_{x\to0}\frac{\cos(\alpha x)-\cos(0)}{x-0}\frac{\sin(x)-\sin(0)}{x-0}=\alpha(-\alpha\sin(0))\cos(0)=0,\\
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\lim_{x\to0}\frac{\sin(\alpha x)\big(\cos(x)-\cos(0)\big)}{(x-0)^2}&amp;=\lim_{x\to0}\frac{\sin(\alpha x)-\sin(0)}{x-0}\frac{\cos(x)-\cos(0)}{x-0}=\alpha\cos(0)(-\sin(0))=0
\end{align*}
by definition of the derivative at 0. Next,
\begin{align*}
\lim_{x\to0}\frac{\alpha\cos(0)\sin(x)-\sin(\alpha x)\cos(0)}{(x-0)^2}&amp;=\cos(0)\lim_{x\to0}\frac{\alpha\sin(x)-\sin(\alpha x)}{(x-0)^2}\\
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&amp;=\cos(0)\lim_{x\to0}\frac{\left[\frac{\alpha\big(\sin(x)-\sin(0)\big)}{x-0}-\alpha\cos(0)\right]-\left[\frac{\sin(\alpha x)-\sin(0)}{x-0}-\alpha\cos(0)\right]}{x-0}\\
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&amp;=\cos(0)\big(-\alpha\sin(0)+\alpha\sin(0)\big)\\
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&amp;=0
\end{align*}
by definition of the second derivative at 0. As the limit of products is the products of the limits (provided that all limits exist in $\mathbb{R}$), we therefore have established the following:
\begin{align*}
\lim_{x\to0}f'(x)&amp;=0.
\end{align*}
On the other hand,
\begin{align*}
\lim_{x\to0}\frac{f(x)-f(0)}{x-0}&amp;=\lim_{x\to0}\frac{\frac{\sin(\alpha x)}{\sin(x)}-\alpha}{x-0}\\
%
&amp;=\lim_{x\to0}\frac{x-0}{\sin(x)}\frac{\sin(\alpha x)-\alpha\sin(x)}{(x-0)^2}\\
%
&amp;=\lim_{x\to0}\frac{x-0}{\sin(x)-\sin(0)}\frac{\sin(\alpha x)-\alpha \sin(x)}{(x-0)^2}
\end{align*}
is proved to be 0 as well using the above computations. This shows that $f'(0)=0$ and thus $f'$ is continuous at $x=0$.
Remark. The above proof works as well for any function $f(x)=\frac{g(\alpha x)}{g(x)}$ where $g$ is smooth and satisfies $g(0)=0$ and $g'(0)=1$.