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Let $f: \mathbb{R}^n \rightarrow \mathbb{R}$ and $g: \mathbb{R}^n \rightarrow \mathbb{R}$ be differentiable. Show the following limit is true:

\begin{align*} \lim_{h \rightarrow 0} \frac{|(f(x+h) - f(x))(g(x+h) - g(x))|}{\|h\|} = 0. \end{align*}

I believe the approach is to use the fact that $f$ and $g$ are continuous to show that $\lim_{h \rightarrow 0} f(x+h) - f(x) = 0$, and analogously for $g$, but I'm confused by the denominator. Is it possible to invoke continuity without doing more manipulations of the quotient?

Note that this limit is part of the proof for the multivariable product rule.

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    $\begingroup$ Hint:$$\frac{|(f(x + h) - f(x))(g(x + h) - g(x))|}{\|h\|} = \|h\|\left|\frac{f(x + h) - f(x)}{\|h\|}\right|\left|\frac{g(x + h) - g(x)}{\|h\|}\right|.$$ $\endgroup$ – Theo Bendit Feb 14 at 18:17
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    $\begingroup$ You should use differentiability and the fact that a limit of product is a product of limits (when both exist). $\endgroup$ – Timur Bakiev Feb 14 at 18:17
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You know that $g$ is differentiable, so you know that $$ \frac{|g(x+h)-g(x)|}{\|h\|}=\frac{|\sum_{i=1}^n\frac{\partial g}{\partial x_i}(x)h_i+o(\|h\|)|}{\|h\|}\le \left|\sum_{i=1}^n\frac{\partial g}{\partial x_i}(x) \right| + \varepsilon(h) $$ where $\varepsilon(h)\to 0$ as $h\to 0$. So the second factor in your limit is bounded while the first, as you indicated, approaches zero. Therefore the limit is zero.

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    $\begingroup$ Apparently to many users the difference between $||h||$ and $\|h\|$ is not conspicuous. But look at $||h|| ||k||,$ coded as ||h|| ||k||, versus $\|h\|\|k\|,$ coded as \|h\|\|k\|. $$ \begin{align} & {} \\ & ||h|| ||k|| \\ {} \\ & \|h\|\|k\| \end{align} $$ $\endgroup$ – Michael Hardy Feb 14 at 18:58
  • $\begingroup$ @MichaelHardy You are right. I was just being lazy. $\endgroup$ – GReyes Feb 14 at 19:03

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