Show that $e^{x+2y}$ is differentiable in $(2,2)$ using the definition of differentiability I kind of understand the rough technique, but the above example was to hard for me. I understand that to show that a function $f(x,y)$ is differentiable at point $(a,b)$ I have to show:
$$
\lim_{(h,k) \to (0,0)}\frac{f(a+h,b+k)-f(a,b)-f_x'(a,b)h-f_y'(a,b)k}{\sqrt{h^2+k^2}}=0
$$
So with $(a,b)=(2,2)$ and $f(x,y)=e^{x+2y}$, I get:
$$
e^6\lim_{(h,k) \to (0,0)}\frac{e^{h+2k}-1-(h+2k)}{\sqrt{h^2+k^2}}
$$
Several uninteresting intermediate calculations have been omitted. Like $f_x'(2,2)=e^6$ and $f_y'(2,2)=2e^6$. 
I don't know how to continue here. I have attempted the substitution $x=h+2k$, $y=h-2k$ which gives me:
$$
4e^6\lim_{(x,y)\to(0,0)}\frac{e^x-1-x}{\sqrt{5x^2+5y^2+6xy}}$$
In P.F:
$$
4e^6\lim_{r \to 0}\frac{e^{r\cos\theta}-r\cos\theta-1}{\sqrt{5r^2+6r^2\cos\theta\sin\theta}}
\iff
4e^6\lim_{r \to 0}\frac{e^{r\cos\theta}-r\cos\theta-1}{r\sqrt{5+3\sin2\theta}}
$$
What should I do to proceed?
 A: Note that $|h+2k|\leq\sqrt{5}\sqrt{h^2+k^2}$, by Schwarz' inequality. Furthermore, since $\exp$ is differentiable at $6$ with $\exp'(6)=e^6$ there is a function $u\mapsto r(u)$ with $\lim_{u\to0} r(u)=0$ such that
$$e^{6+u}=e^6+e^6\, u+ u\, r(u)\qquad(u\to0)\ .\tag{1}$$
Let $f(x,y):=e^{x+2y}$. Putting $u:=h+2k$ in $(1)$ then gives
$$e^{(2+h)+2(2+k)}=e^{6+h+2k}=e^6+e^6(h+2k)+(h+2k)r(h+2k)\ .$$
It follows that
$${f(2+h,2+k)-f(2,2)-e^6(h+2k)\over\sqrt{h^2+k^2}}={h+2k\over\sqrt{h^2+k^2}}r(h+2k)\ .$$
Here the first factor on the RHS is $\leq\sqrt{5}$ in absolute value, and the second converges to $0$ when $(h,k)\to(0,0)$. This proves that $f$ is differentiable at $(2,2)$, and that $\nabla f(2,2)=(e^6,2e^6)$.
Note that the differentiability of $f$ follows from "general principles", since $f=\exp\circ\,\psi$, whereby both $\exp$ and $\psi:\>(x,y)\mapsto x+2y$ are differentiable.
A: How about if you stop at the expression just above where you said "Several uninteresting..." and put h and k in polar form, i.e. $h=r\cos\theta$ and $k=r\sin\theta$ and also make the substitution $p=h+2k$ 
The limit then is 
$$\lim_{p \to 0} e^6 (\cos\theta + 2\sin\theta)  \frac{e^p-(1+p)}{p}$$
But $$\lim_{p \to 0} \frac{e^p-(1+p)}{p} = 0$$ (property of exponential function that you can prove by looking at the series expansion of the exponential function, for example).
and the expression involving thetas in parentheses is bounded.  So you have that the limit is zero.
