directional derivative of $f(x,y,z) = e^{xyz}, v = (1,1,-1)$ through differential quotient (aka no Jacobian)

Question

I want to calculate the directional derivative of $f(x,y,z) = e^{xyz}, v = (1,1,-1)$ without recourse to the Jacobian matrix (in which case this is a very easy exercise). Here's what I've got so far: $$\frac{\partial}{\partial_v}f=\lim_{h \to 0} \frac{e^{(x+h)(y+h)(z-h)}-e^{xyz}}{h}=e^{xyz} \cdot \lim_{h \to 0} \frac{e^{h(-xy+xz+yz)+h^2(-x-y+z)-h^3}-1}{h}.$$ The answer is now intuitively clear, i.e. the higher order $h$ terms go to zero faster than the linear $h$ term, thus I'm left with$$ e^{xyz} \cdot \lim_{h \to 0} \frac{e^{h(-xy+xz+yz)}-1}{h} = e^{xyz} (-xy+xz+yz). $$ But two things to be made precise:

$1.$ why is $\lim_{h \to 0} \frac{e^{hk}-1}{h} =k $

$2.$ why exactly can I let the higher order $h$ terms in the exponent go to zero?

Answer 1

By definition the directional derivative you are after is the quantity $${d\over dh}e^{(x+h)(y+h)(z-h)}\biggr|_{h=0}\ .$$ Now $${d\over dh}\bigl((x+h)(y+h)(z-h)\bigr)=(y+h)(z-h)+(x+h)(z-h)-(x+h)(y+h)\ ,$$ and letting $h=0$ here gives $yz+xz-xy$ on the RHS. It follows that $${d\over dh}e^{(x+h)(y+h)(z-h)}\biggr|_{h=0}=(yz+xz-xy)e^{xyz}\ .$$ Concerning your last questions:

1. When $k=0$ this is trivially true. When $k\ne0$ you can write $${e^{kh}-1\over h}=k\cdot{e^{kh}-1\over kh}\to k\cdot1\qquad(h\to0)\ .$$
2. You cannot. But this is a useful heuristic device which in many cases leads to a correct result. Experience is needed if you don't want to fall into a trap.

Answer 2

The taylor series expansion for $e^x$ is:- $$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}...$$

Thus $$\frac{e^x-1}{x}=1+\frac{x}{2!}+\frac{x^2}{3!}+\frac{x^3}{4!}...$$

Hence the limit is $1$

ANOTHER EXPLANATION :-

$$\lim_{x\to0}\frac{e^x-1}{x}=\lim_{x\to0}\frac{e^{x}-e^0}{x-0}=\Big[\frac{d}{dx}e^x\Big]_{x=0}=\Big[e^x\Big]_{x=0}=1$$