# Rigorous proof for a limit of a function

Consider the following expression: $$\lim_{(x,y) \to (0,0)}g(x,y)=\ln(1+3x+4y+x^2+y^4)$$

I have learned in the past that if a function is continuous, I can simply substitute the points into the function, and obtain a result -- in this case, that would be $\ln(1)=0$. However, this does not seem sufficient enough to me, as we are making this base assumption of continuity. Does this require more rigor?

Must I prove that $g(x,y)$ is differentiable, which would in turn imply continuity? Must I first prove continuity at $(0,0)$ using the definition? Lastly, how would I prove that this limit is $0$ using an $\varepsilon-\delta$ proof? I am particularly confused by the fact that this is a composite function.

In short, I am looking for suggestions in regards to what would be needed to show that the above limit exists and is equal to $0$. Not looking for a solution, just confused about which particular criteria are needed.

Your function is a composition of two functions, $f:\mathbb{R}^2 \to \mathbb{R}$ defined by $f(x,y) = 1 + 3x + 4y + x^2 + y^4$ and $h: \mathbb{R}_{> 0} \to \mathbb{R}$ defined by $h(z) = \ln z$. You first need to show that near $(0,0)$, your composition of functions is defined (this is not hard, because $f(x,y)$ is close to 1 for $x,y$ close to $0$. Then if you know polynomials are continuous and natural log is continuous (where it is defined), then your composition of functions is continuous (the composition of two continuous functions is continuous) and the limit exists and is simply obtained by plugging $(0,0)$ into your $g$.

The answer ultimately depends on what level of rigor the situation calls for. In a typical multivariable calculus course, for instance, you may usually take for granted that polynomials (in any number of variables) and $f(z) = \ln(z)$ are both continuous functions on their domains, whereas this may or may not be the case in a course on foundational real analysis. So if you're in the former camp (as your tags suggest) then you can probably get away with just stating that the relevant polynomial in $x$ and $y$, $f(z) = \ln z$, and compositions of continuous functions are all continuous.

Now for the fun part: How do you actually show these things with some rigor? Again, that will depend on where you start/how much you can assume, but here are some off-the-cuff answers for how to do this with a reasonable toolset at your disposal. I've left the $(\epsilon,\delta)$ arguments in a form that can probably be made cleaner but that gives you an idea of how to cook these sorts of proofs up yourself.

# Claim: Every polynomial $P(x,y)$ is continuous on $\mathbb{R}^2$.

Proof: Note first that $q_C(x,y) = C$ for any constant $C$, $P_1(x,y) = x$, and $P_2(x,y) = y$ are all continuous functions. (Using the usual $(\epsilon,\delta)$ definition of continuity at a point, given $\epsilon > 0$ set $\delta = \epsilon$.) Since products and sums of continuous functions are continuous (by the same argument used in one variable) and $P(x,y)$ is a sum of products of functions of the form $q_C$ (for varying $C$), $P_1$, and $P_2$), it follows that $P$ is continuous.

# Claim: $f(z) = \ln(z)$ is continuous at every $z > 0$

Proof: The usual formal definition of $\ln(z)$ for $z > 0$ is

$$\ln(z) = \int_1^z \frac{1}{t}\,dt.$$

Since $1/t$ is defined and Riemann integrable on $[1,z]$ (or $[z,1]$ for $0<z<1$) one can show that $\ln(z)$ is continuous.

# Claim: Compositions of functions are continuous

Suppose $A \subset \mathbb{R}^m$, $B \subset \mathbb{R}^n$, and $C \subset \mathbb{R}^p$. Suppose $f: A \to \mathbb{R}^n$ is continuous at $a$ with $f(a) \in B$ and $g: B \to C$ is continuous at $f(a)$. Then we must show that $h = g \circ f$ is continuous at $a$.

Let $\epsilon > 0$. Since $g$ is continuous at $f(a)$, there exists $\delta_1 = \delta_1(\epsilon) > 0$ such that if $y \in B$ and $\|y-f(a)\| < \delta_1$ then $\|g(y) - f(a)\| < \epsilon$. When considering $f$, think of $\delta_1$ as taking the role of your given $\epsilon$. Since $f$ is continuous at $a$, there exists $\delta > 0$ such that if $x \in A$ and $\|x-a\| < \delta$ then $\|f(x) - f(a)\| < \delta_1$. Thus for all $x \in A$ with $\|x-a\| < \delta$ we have that $f(x)$ is a point in $B$ at which $\|f(x)-f(a)\| < \delta_1$, and so by the above statement about the continuity of $g$ at $f(a)$ we conclude $\|h(x) - h(a)\| = \|g(f(x)) - g(f(a))\| < \epsilon$.