Prove that $f$ is continuous at $(0, y_0)$ where f is defined on $\mathbb{R}^2$ as follows:

Question

Prove that $f$ is continuous at $(0, y_0)$ where $f$ is defined on $\mathbb{R}^2$ as follows:

$$f(x, y) = \begin{cases} (1+xy)^{1/x} &\mbox{if } x \neq 0 \\ e^y & \mbox{if } x = 0 \end{cases}$$

Attempt:

My approach was to show that this function is a composition of continuous functions. In particular the functions: $$g(x) = (1+x)$$ $$h(x) = xc \text{ (where $c$ is a constant)}$$ $$n(x) = c^{\frac{1}{x}} $$

Now I have successfully shown that $g(x)$ and $h(x)$ are continuous at my point, but I am having problems with $n(x)$ and $e^y$. I think $e^y$ is constant at this point and as such I could just let $\delta = \epsilon$. But $n(x)$ definitely has me confused.

Answer 1

You can use the well-known limit $\lim_{x \to 0} (1+kx)^{\frac 1x} = e^k$. Hence we have:

$$\lim_{(x,y) \to (0,y_0)} f(x) = \lim_{(x,y) \to (0,y_0)} (1+xy)^{\frac 1x} = \lim_{(x,y) \to (0,y_0)} e^y = e^{y_0} = f(0,y_0)$$

Hence the function is continuous at $(0,y_0)$.

Answer 2

It is enough to show that $lim_{(x,y)\rightarrow (0,y_0), y\neq 0}f(x,y)=f(0,y_0)$.

$f(x,y)=e^{{ln(1+xy)}\over x}$ if $x\neq 0$, $lim_{(x,y)\rightarrow (0,y_0})e^{{ln(1+xy)}\over x}$

$lim_{x\rightarrow 0}e^{{ln(1+xy_0)}\over x}=e^{y_0}$

since $lim_{x\rightarrow 0}{{ln(1+xy_0)}\over x}$ is the derivative of $ln(1+xy_0)$ at $0$.