I want to see whether the Cobb-Douglas function $f(x,y) = x^\alpha y^\beta$ is differentiable at $(x,y)=(0,0)$. Both partial derivatives are zero $f^\prime_x(0,0) = f^\prime_y(0,0)= 0$. I need to check if the following limit is zero $$ \lim_{(x,y)\rightarrow (0,0)} \frac{f(x,y)-f(0,0) - f^\prime_x(0,0) x - f^\prime_y(0,0) y}{\sqrt{x^2 + y^2}} = 0 $$ then the function is differentiable at the origin.
Replacing I get $$ \lim_{(x,y)\rightarrow (0,0)} \frac{x^\alpha y^\beta}{\sqrt{x^2 + y^2}} $$
If $g(x,y) = \frac{x^\alpha y^\beta}{\sqrt{x^2 + y^2}}$ then $$ \lim_{t \rightarrow 0} g(t,0)= \lim_{t \rightarrow 0} \frac{0}{t} = 0 $$ But $$ \lim_{t \rightarrow 0} g(t,t)= \frac{1}{\sqrt{2}} \lim_{t \rightarrow 0} t^{\alpha+\beta-1} $$
And this last expression is not defined for $\alpha+\beta=1$, and is not equal to zero for $\alpha+\beta<1$
Is this okay? How about $\alpha+\beta>1$? Can someone help me prove that it is not differentiable?
You are correct that $\lim_{t\to0} t^{\alpha + \beta - 1}$ does not exist if $\alpha + \beta < 1$ (the limit is $+\infty$). So $f$ is not differentiable if $\alpha + \beta < 1$.
You are not correct in the case that $\alpha+\beta = 1$. In this case $t^{\alpha + \beta - 1} = 1$, so the limit exists. But it's not equal to zero, which means $f$ is not differentiable at $(0,0)$.
Now if $\alpha + \beta > 1$, then $\lim_{t\to 0} \frac{1}{\sqrt{2}} t^{\alpha + \beta - 1} = 0$. That isn't enough to say that $\lim_{(x,y) \to (0,0)} \frac{x^\alpha y^\beta}{\sqrt{x^2+y^2}} = 0$, however.
If $\alpha + \beta > 1$, we can prove that $f$ is differentiable at $(0,0)$ with the squeeze theorem. Note that for all real $x$ and $y$, $$ 0 \leq |x| \leq \sqrt{x^2+y^2} \qquad \text{and} \qquad 0 \leq |y| \leq \sqrt{x^2+y^2} $$ This means that if $(x,y) \neq (0,0)$, $$\begin{aligned} 0 &\leq \left| \frac{x^\alpha y^\beta}{\sqrt{x^2+y^2}} \right| = \frac{|x|^\alpha |y|^\beta}{\sqrt{x^2+y^2}} \\&\leq \frac{(x^2+y^2)^{\alpha/2}(x^2+y^2)^{\beta/2}}{(x^2+y^2)^{1/2}} \\&= (x^2+y^2)^{(\alpha + \beta -1)/2} \end{aligned}$$ Since $\lim_{(x,y) \to (0,0)} (x^2+y^2)^{(\alpha + \beta -1)/2} = 0$, the term in between must also tend to zero.