Finding whether a parametric curve has a well defined tangent at the origin 
Does the curve $$ x(t) = 5t^3, y(t) = 4t - sin(4t), t \in \Bbb{R}$$ Have a well defined tangent line at the origin?

Since $\frac{dy}{dx} = \frac{0}{0} $ When t = 0 (i.e at the origin) my guess is that the curve does not have a well defined tangent at the origin, however this is not correct apparently.
 A: The curve has a very well defined tangent. 
EDIT1
(After hasty post!)
By L'Hospital's Rule applied twice due to $\frac{0}{0}$ persisting after first application the Rule
$$ \frac{dy}{dx}=\frac{\dot y}{\dot x}= \frac{4-4 \cos \,t}{15\, t^2} =\frac{4}{15} \frac{1-\cos 4t}{t^2}=\frac{4}{15} \frac{4 \sin 4t}{2t} \ =
\frac{4}{15} \cdot 2\cdot 4 =\frac{32}{15} $$
is the slope of tangent at origin with a second order contact.
A: $\frac {dy}{dx} =  \frac {\frac {dy}{dt}}{\frac {dx}{dt}} = \lim_\limits{t\to 0}\frac {4(1-\cos 4t)}{15t^2} = \frac {64}{30}$
$t = \left(\frac {x}{5}\right)^{\frac 13}\\
y = 4\left(\frac {x}{5}\right)^{\frac 13} - \sin (4\left(\frac {x}{5}\right)^{\frac 13})\\
\frac {dy}{dx} = \lim_\limits{h\to 0} \frac{y(h)-y(0)}{h}$ 
Which is defined and indeed equals $\frac {32}{15}$
A: use that $$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$
A: $\frac{dy}{dx}=\frac{y'}{x'}=\frac{0}{0}$, You can actually compute the limit of the tangent lines: $\begin{align}
\lim_{t \to 0}\frac{y'(t)}{x'(t)}=\\
=\lim_{t \to 0} \frac{4-4\cos(4t)}{30t^2}=\\
=\lim_{t \to 0} \frac{16}{15}\frac{4\left(1-\cos(4t)\right)}{\left(4t\right)^2}=\\
=\lim_{t \to 0}\frac{\sin(4t)}{30t}=\\
=\frac{64}{30}=\frac{32}{15}
\end{align}$ 
If you are not comfortable with The L'Hospital's theorem, consider directly 
$\begin{align}
\lim_{x \to 0}\frac{1-\cos(x)}{x^2}=\\
=\lim_{x \to 0}\frac{2\sin^2\left(\frac{x}{2}\right)}{x^2}=
\\=\lim_{\frac{x}{2} \to 0}\frac{2\sin\left(\frac{x}{2}\right)}{4\left(\frac{x}{2}\right)^2}=\\
=\frac{1}{2}
\end{align}$
And apply it to the third passage of the computation above with the substitution $4t=z$
Actually, an equivalent (even if a bit lengthy) computation might be carried out with $\begin{align}\lim_{x \to 0}y'(x)=\\
=\lim_{x \to 0}\frac{d}{dx}\left(4\left(\frac{x}{5}\right)^{\frac{1}{3}}-4\sin\left(\left(\frac{x}{5}\right)^{\frac{1}{3}}\right)\right)=\\
=\lim_{h \to 0}\frac{\left(4\left(\frac{h}{5}\right)^{\frac{1}{3}}-4\sin\left(\left(\frac{h}{5}\right)^{\frac{1}{3}}\right)\right)}{h}=\\
=\frac{32}{15}
\end{align}$,
(the term $y(0)$ vanishes since it equals 0, and the result is obtained applying The L'Hospital theorem or, equivalently, substituting back $5t^3=x$)
A: Using the equivalence $1-\cos{t} \sim \frac{t^2}{2}, \ t\to{0},$ we have 
$$y'(0) = \lim\limits_{t\to{0}}\dfrac{y'(t)}{x'(t)} = \lim\limits_{t\to{0}}\dfrac{4-4 \cos{4t}}{15t^2} = \lim\limits_{t\to{0}}\dfrac{32t^2}{15t^2} = \dfrac{32}{15}.$$
