Gradient of smooth function on unit sphere near the north pole

The two unit vectors

$$\left\{\frac{(-y,x,0)}{\sqrt{x^2+y^2}}, \frac{(-xz,-yz,x^2+y^2)}{\sqrt{x^2+y^2}}\right\}$$

form an orthonormal basis to the tangent space at $(x,y,z)\in \mathbb{S^2}\backslash {(0,0,\pm1)}$. Observe that $$(-xz,-yz,x^2+y^2)=(x,y,z)\times (-y,x,0).$$

Now let $u$ be a smooth function on $\mathbb{S}^2$ and define

$$\nabla u (x,y,z) \cdot \frac{(-xz,-yz,x^2+y^2)}{\sqrt{x^2+y^2}}=\nabla u (x,y,z) \cdot \frac{(-xz,-yz,1-z^2)}{\sqrt{x^2+y^2}}=\nabla u (x,y,z) \cdot \frac{(0,0,1)}{\sqrt{x^2+y^2}},$$where I have used $\nabla u(x,y,z)\cdot (x,y,z)=0$.

Since $\nabla u (0,0,1)$ is parallel to $xy-$plane, intuitively one expect the projection of $\nabla u$ onto $\frac{(-xz,-yz,x^2+y^2)}{\sqrt{x^2+y^2}}$ to be small as $(x,y,z)\rightarrow 0$, and hence$$\lim_{(x,y,z)\rightarrow (0,0,1)}\nabla u (x,y,z) \cdot \frac{(0,0,1)}{\sqrt{x^2+y^2}}=0.$$

Is the above limit zero? I am unable to prove or disprove it. I would appreciate any hint.