# Taylor Series expansion when derivative is 0/0

I have a tutorial question on Taylor series.

$$f(x,y,z) = e^{(\sqrt{x y}+\sqrt{x z}+\sqrt{z y})}$$

I managed to get the Taylor series at (1,1,1)

Is it possible to get the expansion at (0,0,0)? Because at (0,0,0) the derivative is 0/0 which is confusing...how do we evaluate 0/0 in Taylor series?

Thanks!

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$f$ is not differentiable at $0$, so there is no Taylor series... – copper.hat Aug 30 '12 at 5:31
Would it be possible to make approximation for small x? – Ken Ryou Aug 30 '12 at 5:42
Well, it depends on what you mean, since $e^x \approx 1+x$ near $0$, you could approximate $f(x) \approx 1+\sqrt{x y}+\sqrt{x z}+\sqrt{z y}$. – copper.hat Aug 30 '12 at 5:45
I just read about L hospital, would it be applicable in this case? Thanks! – Ken Ryou Aug 30 '12 at 6:59
Not really, $f$ is really not differentiable at $0$. Loosely, the problem is that $x \mapsto \sqrt{x}$ has infinite slope at 0. (If it has a Taylor series, it would be differentiable; that's why $f$ has no Taylor series at $0$.) – copper.hat Aug 30 '12 at 7:01

As copper.hat said in the comments, it is not possible to do a Taylor series in $x$, $y$ or $z$ around the point $(0,0,0)$ because the function has no derivative in that point.
What you can do in this point, however, is to do a Taylor expansion of just the exponential function, and insert the argument; this will give you an expansion in square roots: \begin{aligned} \mathrm e^{\sqrt{xy}+\sqrt{xz}+\sqrt{yz}} &= \sum_{n=0}^\infty \frac{1}{n!}(\sqrt{xy}+\sqrt{xz}+\sqrt{yz})^n\\ &= 1 + \sqrt{xy}+\sqrt{xz}+\sqrt{yz}\\ &\phantom{= 1} + \frac12 xy+\frac12 xz+\frac12 yz + \sqrt{xy}\sqrt{xz}+\sqrt{xy}\sqrt{yz}+\sqrt{xz}\sqrt{yz} + \dots \end{aligned}