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Relevant wiki: Derivative by First Principle

My solution is similar to Brian's:

Substitute $x^2+y^2+z^2=t$ and note that $\lim_{t\to 0^+}\frac{e^{-t}-1}{t}=\boxed{-1}$ , the derivative of $e^{-t}$ at $t=0$ .

Applying spherical coordinates, let $x = r\sin(\theta)\cos(\phi), y = r\sin(\theta)\sin(\phi), z = r\cos(\theta)$ , where $r \in (0,\infty), \theta \in [0, \pi], \phi \in [0,2\pi)$ . Then $x^{2} + y^{2} + z^{2} = r^{2}$ , and the three-variable limit is transformed into a one-variable limit to which L'Hopital's rule can be applied as follows:

$\large\displaystyle\lim_{r \to 0} \dfrac{e^{-r^{2}} - 1}{r^{2}} = \lim_{r \to 0} \dfrac{-2re^{-r^{2}}}{2r} = \lim_{r \to 0} -e^{-r^{2}} = \boxed{-1}$ .

Substituting $x^{2}+y^{2}+z^{2} = u$ , we notice that as $(x,y,z)→(0,0,0)$ , $u→0$

Next, we transform the limit to read $\lim_{u\to0} \frac{e^{-u}-1}{u}$

Since the value of the numerator and denominator at $u=0$ is $0$ , we can apply l'Hôpital's Rule

$\lim_{u\to0} \frac{e^{-u}-1}{u} = \lim_{u\to0} \frac{\frac{d}{du}(e^{-u}-1)}{\frac{d}{du}u} = \lim_{u\to0} \frac{-e^{-u}}{1} = \lim_{u\to0} -e^{-u} = -(e^{0}) = -(1) = \boxed{-1}$

The fastest would be to consider the series expansion for $\large e^x$ . :-)