Intersection of line/manifold converges if it converges for line/tangent space? Let $f : \mathbb R^n → \mathbb R$ be a continuously differentiable function with non-zero gradient. Then, according to the implicit function theorem a level set defines a $n-1$ dimensional manifold $M$ and the tangent space is well defined.
Now, assume that there is a sequence of lines, each intersecting the tangent space (at a point $x_0$) at point $x_m$ with an angle larger than $\varphi_0$, and the limit 
$\hspace{1cm}\lim_{m\to\infty}x_m=x_0, \hspace{1cm}\text{ with } m\in\mathbb{N}, x_0 \in M.$
In general, $x_m \notin M$.
Does the intersection of the line and $M$ converge towards $x_0$?
 A: $\newcommand{\Reals}{\mathbf{R}}$In a word, yes. Here's a sketch.
By renaming coordinates if necessary, the implicit function theorem allows you to assume $x_{0}$ is the origin, that $y_{n} = g(y_{1}, \dots, y_{n-1})$ for some $C^{1}$ function $g$ defined in a neighborhood of the origin in $\Reals^{n-1}$, and that the tangent space to $M$ at $x_{0}$ is the hyperplane $\{y_{n} = 0\}$.
By hypothesis, there is a sequence of lines given parametrically by $\ell_{m}(t) = x_{m} + t v_{m}$, with $v_{m}$ a unit vector in $\Reals^{n}$, and with the $n$th coordinates of the $v_{m}$ greater than $M := \sin(\varphi_{0}) > 0$ in absolute value.
Consider the (two-napped) cone
$$
y_{n} = \pm\tfrac{1}{2}M(y_{1}^{2} + \dots + y_{n-1}^{2})^{1/2} := \pm h(y).
\tag{1}
$$
Because $g$ is of class $C^{1}$ and is tangent to $\{y_{n} = 0\}$ at $0$, there exists an open ball $U$ in $\Reals^{n-1}$ such that $|g(y)| \leq h(y)$ for all $y$ in $U$.
It's now straightforward to calculate (or estimate) where $\ell_{m}$ intersects the cone (1), to show the points of intersection approach the origin, and consequently to see that $\ell_{m}$ intersects $M$ somewhere on a segment whose endpoints approach $0$, so the points of intersection of $M$ and the lines $\ell_{m}$ approach the origin as well.
