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Let $f(x,y) \rightarrow R$ be a function with continuous partial derivatives, and let $S$ be the surface $z=f(x,y)$ in $R^2$. Let $P_{0}=(x_0,y_0,z_0 )$ be a point in S and $P=(x,y,z)$ be some other point in S. We're asked to show that $a$, the angle between the plane tangent to S at $(x_0,y_0,z_0)$ and the vector $P-P_0$, approaches $0$ as $P\rightarrow P_0$.

I'd appreciate some help with proving this. The question is from a former exam in my multivariable calculus course.

The angle thing ticks me off here since I have no idea how to approach this. I tried proving this using dot products to derive the angle between the vectors, but calculating the limit was difficult.

Thanks!

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Hint only, since this looks like homework and you should try to do the calculations yourself:

Instead of $P-P_0$, consider the corresponding normalized unit vector $Q= \frac{P-P_0}{|P-P_0|}$, so you have not to bother about estimating it's length. Instead of the tangent plane, consider the unit normal $n$ to the tangent plane and show that $\langle n,Q\rangle$ tends to zero as $P$ approaches $P_0$.

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    @ro44 that's the general idea with hints ;-) Good luck then. (For exam preparation it is, compared to homework, even more helpful for you if you do most of the proofs yourself.)2012-08-19
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$P-P_0=(x_0-x\,,\,y_0-y\,,\,z_0-z)$

As a plane's normal at $\,P\,$ is $\,N:=\left(f_x(P)\,,\,f_y(P)\,,\,-1\right)\,$ , we get that the sine of the angle between the line and the plane is $(P-P_0)\cdot N=f_x(P)(x_0-x)+f_y(P)(y_0-y)-(z_0-z)\xrightarrow [(x,y,z)\to(x_0,y_0,z_0)]{}0$ since the partial derivatives $\,f_x\,,\,f_y\,$ are given continuous at $\,P\,$