Finding limits for multivariable functions

Question

I am new to limits of multivariable functions. One of the first thing I learnt is that one of the ways we can evaluate a limit is by plugging in the values so long as it is continuous as per the theorem below.

I was doing a tutorial question and after I compared my solution to the solution from my professor (pictured below), it seems to me that we find that the function is continuous first and then plug in the values to find the limit.

Why didn’t we have to prove continuity by showing the limit itself equals the function at f(a, b) as the theorem would suggest and why did we just say that since it is continuous on that domain then we can plug it in? How can we say it’s continuous if we didn’t prove it? Does it have to do with the fact that continuity from functions in single variables extends to the same functions which are multivariable?

I hope someone can clarify this for me.

Answer 1

The key theorems your professor (implicitly) used are these two: $\newcommand{\real}{\mathbb{R}}$

Let $f,g$ be continuous functions where the range of $f$ is inside the domain of $g$. Then $g\circ f$ is continuous.

Let $f,g:A\to\real$ be continuous functions, where $A\subseteq\real^n$. Then $f+g$ and $fg$ are continuous.

I'll leave the proofs to you as a good beginner exercise (not much difference than the 1D case!). Now, back to your function $f(x,y)=\exp(\sqrt{2x-y})$. We can deconstruct it step by step.

1. The function $\real^2\to\real$ defined by $(x,y)\mapsto x$ is continuous. The same is true for $(x,y)\mapsto y$. The constant functions $(x,y)\to k$ for any $k\in\real$ are also continuous. These can be proved from definitions.
2. By the second theorem above, $(x,y)\mapsto 2x-y$ is continuous.
3. The function $\sqrt{\cdot}:\real_{\ge 0}\to\real$ is continuous, so $(x,y)\mapsto\sqrt{2x-y}$ is continuous, by the first theorem. (We assume the domain is such that $2x-y\ge 0$, or else the function is undefined.)
4. The function $\exp:\real\to\real$ is continuous, so $(x,y)\mapsto\exp(\sqrt{2x-y})$ is continuous, by the first theorem. This shows that $f$ is continuous.

Finally, we use the following theorem:

Let $f:A\to\real$ be a continuous function, where $A\subseteq\real^n$. Let $a\in A$. If $a$ is a limit point of $A$, then $$ \lim_{x\to a} f(x)=f(a).$$

This allows us to "plug in" values into a continuous function to compute its limit at a certain point. In your case, we have $\displaystyle\lim_{(x,y)\to(3,2)} f(x,y)=f(3,2)=e^2$.