2 equivalent definitions of differentiability

Question

I learned in class that to prove by definition the differentiability in(a,b) of a function $f$ we need to show that $f$ can be presented as :

(*)$f(k+a,t+b) -f(a,b)-f_x(a,b)k-f_y(a,b)t = \epsilon(k,t)\sqrt{k^2 +t^2}$ for $\epsilon \rightarrow0$ as $k,t \rightarrow 0$

but then I saw this definition in one of the solutions (it was presented as definition):

(**) $f$ is differentiable if: $\lim_{k,t\rightarrow0}\frac{f(k,t) -f(a,b)-f_x(a,b)k-f_y(a,b)t}{\sqrt{k^2 +t^2}} = 0$

I do understand that $(*)\Rightarrow (**)$, but I am not sure if I am correct about the other way:

Is it because $\epsilon$ epsilon is $\frac{f(k,t) -f(a,b)-f_x(a,b)k-f_y(a,b)t}{\sqrt{k^2 +t^2}}$?

And if not, why does $(**)\Rightarrow (*)$

Answer 1

Exactly. As you said, simply choose

$$\epsilon(k,t)=\frac{f(k,t) -f(a,b)-f_x(a,b)k-f_y(a,b)t}{\sqrt{k^2 +t^2}}$$

for $(k,t)\not=(0,0)$, which then satisfies your version $(*)$.

$$\lim_{k,t\to 0} \epsilon(k,t)=\lim_{k,t\to 0} \frac{f(k,t) -f(a,b)-f_x(a,b)k-f_y(a,b)t}{\sqrt{k^2 +t^2}}=0.$$

Answer 2

If $(**)$, then you define $\varepsilon(x,y)$ as$$\begin{cases}\dfrac{f(k,t) -f(a,b)-f_x(a,b)k-f_y(a,b)t}{\sqrt{k^2 +t^2}}&\text{ if }(k,t)\neq(0,0)\\0&\text{ otherwise.}\end{cases}$$and then $\lim_{(k,t)\to(0,0)}\varepsilon(k,t)=0$ and $(*)$ holds.