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Suppose that the continuous function $f: \Bbb R^2 \to \Bbb R$ has a tangent plane at the point $(0, 0, f(0, 0)).$ Prove that the function $f$ has directional derivatives in all directions at the point $(0, 0).$

My attempt: The tangent plane at the point $(0, 0)$ has the form $z=ax+by+f(0,0)$ with the property $$\lim_{(x,y)\rightarrow(0,0)}\frac{f(x,y)-ax-by-f(0,0)}{\sqrt{x^2+y^2}}=0 \ \ (1).$$

The directional derivative of $f$ in the direction $p=(c,d)$ at $(0,0)$ is $$\frac{\partial f}{\partial p}(0,0)=\lim_{t\rightarrow 0}\frac{f(tc, td)-f(0,0)}{t} \ \ (2).$$

How can I use $(1)$ to prove that $(2)$ exists?

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Your limit $(1)$ is valid for any path of approach towards the origin. So choose the path $y = \frac{d}{c}x$, and plug that in:

$$ 0 = \lim_{x \to 0} \frac{f(x,\frac{d}{c}x) - \left(a+ \frac{bd}{c}\right)x - f(0,0)}{\frac{x}{c} \sqrt{c^2+d^2}} = \lim_{x \to 0} \frac{f(x,\frac{d}{c}x)-f(0,0)}{\frac{x}{c}\sqrt{c^2+d^2}} - \frac{ ac + bd }{\sqrt{c^2+d^2}} $$

Move the negative term to the other side, and cancel the $\sqrt{c^2+d^2}$ factor:

$$ ac+bd = \lim_{x \to 0} \frac{f(x,\frac{d}{c}x) - f(0,0)}{x/c} $$

Just re-parameterize $t = \frac{x}{c}$, and this becomes

$$ ac + bd = \lim_{t \to 0} \frac{f(ct,dt)-f(0,0)}{t} $$

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  • $\begingroup$ where did the denominator $\sqrt{x^2+y^2}$ go? $\endgroup$ – dxdydz Feb 9 at 4:54
  • $\begingroup$ Oops, you're right...changing it right now... $\endgroup$ – Nick Feb 10 at 0:38

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