Find the equation of tangent at origin to the curve $y^2=x^2(1+x+x^2)$ How do I find the equation of tangent at $(0,0)$ to the curve $y^2=x^2(1+x+x^2)$ ?
Differentiating and putting the value of $x$ and $y$ gives an indeterminate form. 
Can we trace the curve and geometrically make tangents and find their equation ?
 A: $$y^2=x^2(1+x+x^2)\implies 2yy'=2x(1+x+x^2)+x^2(1+2x)=4x^3+3x^2+2x\implies$$
$$y'(0)=\lim_{x\to 0}\frac{4x^3+3x^2+2x}{2y}=\pm\lim_{x\to0}\frac{4x^2+3x+2}{2\sqrt{1+x+x^2}}=\pm1$$
and thus the tangent doesn't exist since $\;y=|x|\sqrt{1+x+x^2}\;$ and $\;\frac x{|x|}=\pm1\;$
A: $$y^2=x^2+x^3+x^4$$
$$2y\frac{dy}{dx}=2x+3x^2+4x^3$$
$$\implies \frac{dy}{dx}=\frac{2x+3x^2+4x^3}{2y}$$
Now, equation of tangent line at (0,0) is (eqn 1):
$$y=\frac{dy}{dx}\big{|}_{(0,0)}(x)$$
Since $\dfrac{dy}{dx}$ yields 0/0 form, we use limits to evaluate the tangent:
$$\lim_{x,y\to0} \frac{2x+3x^2+4x^3}{2y} $$
$$ \equiv\lim_{x\to0} \frac{2x+3x^2+4x^3}{2|x|\sqrt{1+x+x^2}} $$
Here we find:


*

*$\lim_{x\to0^-} \dfrac{2x+3x^2+4x^3}{2|x|\sqrt{1+x+x^2}}=-1 $ and

*$\lim_{x\to0^+} \dfrac{2x+3x^2+4x^3}{2|x|\sqrt{1+x+x^2}}=+1 $


The limit is known to not exist, as LHL and RHL don't agree. but in my opinion it doesn't matter if this limit of slope function $\dfrac{dy}{dx}$ exists or not, since its a curve.
so from eqn 1
$$\implies y = \pm x$$
