Evaluate the limit $\lim_{x\to 0} \frac{(\tan(x+\pi/4))^{1/3}-1}{\sin(2x)}$ Evaluate the limit
$$\lim_{x\to 0} \frac{(\tan(x+\pi/4))^{1/3}-1}{\sin(2x)}$$
I know the limit is $1\over3$ by looking at the graph of the function, but how can I algebraically show that that is the limit. using this limit: $$\lim_{x \rightarrow 0} \frac{(1+x)^c -1}{x} =c$$? (without L'Hopital Rule)
 A: $$\lim_{x\to 0} \frac{(\tan(x+\pi/4))^{1/3}-1}{\sin2x}=\lim_{x\to 0} \frac{(1+\tan(x+\pi/4)-1)^{1/3}-1}{\tan(x+π/4)−1}\cdot\frac{\tan(x+π/4)−1}{\sin 2x}=$$
$$=\frac{1}{3}\cdot\lim_{x\to 0}\frac{\frac{1+\tan x}{1-\tan x}-1}{\sin 2x}=\frac{1}{3}\cdot\lim_{x\to 0}\frac{2\tan x}{2\sin x\cos x}=\frac{1}{3}\cdot1=\frac{1}{3}$$
A: Use the definition of derivative:  $$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}.$$  Second hint:  multiply numerator and denominator by $x$.
A: $\newcommand{\+}{^{\dagger}}%
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\begin{align}
&\lim_{x \to 0}{\tan^{1/3}(x + \pi/4) - 1 \over \sin\pars{2x}}
=\half\lim_{x \to 0}\bracks{{2x \over \sin\pars{2x}}\,
{\tan^{1/3}(\pi/4 + x) - 1 \over x}}
\\[3mm]&=\half\lim_{x \to 0}\bracks{{\tan^{1/3}(\pi/4 + x) - 1 \over x}}
=
\left.\half\,\totald{\tan^{1/3}\pars{x}}{x}\right\vert_{x\ =\ \pi/4}
=
\half\,\bracks{{1 \over 3}\,\tan^{-2/3}\pars{x}\sec^{2}\pars{x}}_{x\ =\ \pi/4}
\\[3mm]&= {1 \over 3}
\end{align}
$$\color{#0000ff}{\large%
\lim_{x \to 0}{\tan^{1/3}(x + \pi/4) - 1 \over \sin\pars{2x}}
= {1 \over 3}}$$
