I am trying to prove that the vector-valued function $$\vec{R}(t)=e^{-t}\hat{i}+\ln t\hat{j}+\frac{1}{\sqrt{4-t}}\hat{k}$$ is continuous on its domain. I know that its domain is $\mathbb{R}\cup \mathbb{R}^{+}\cup(-\infty,4)$, or basically $(0,4)$.

I am getting rusty in terms of proving that a function is continuous, so please bear with me. As for continuity, as I have learned from our past lessons, there are three conditions.

  1. $\lim_{t\to a}f(t)$ exists for all $t$ on its domain.
  2. $f(a)$ exists for all $t$ on its domain.
  3. $\lim_{t\to a}f(t)=f(a)$.

However, there are parts on which this definition doesn't hold. Let $g(t)=\ln t$. We have $$lim_{t\to 0^{+}}g(t)=lim_{t\to 0^{+}}\ln t=-\infty$$.While this holds for limits, I don't think it will work for $g(0)$.

Despite that, I think it's "reasonable" to say that $g(0)=-\infty$ since the natural logarithm function is not defined for all $t<0$, and it decreases without bound. But we know that it's a mathematical sin to equate infinity to any function.

The same problem is encountered on $\frac{1}{\sqrt{4-t}}$, where $t=4$.

I think I'm doing something wrong here. Perhaps, I'm doing it all wrong. I think there are theorems that explain the continuity of exponential, logarithmic, and rational functions. But I also think our professor wouldn't give a trivial question where we will only need to find the union of the domains of the functions representing the whole of the vector-valued function.

  • $\begingroup$ You are mistaken by the definition of continuity. A function is continuous on its domain $X$ iff $\lim_{x \rightarrow a} f(x) = f(a)$ for all limit points $a \in X$. The $\in X$ is important. $\endgroup$ Apr 29, 2017 at 13:41

1 Answer 1


Vector valued functions are continuous on $(a,b)$ iff each component is continuous on $(a,b)$. Now I show one of the troubling component is continuous on $(0,4)$. You do the rest...

Let $f: (0,4)\rightarrow \Bbb{R}$ be the function defined by $$f(t) =\frac{1}{\sqrt{4-t}}$$ for all $t \in (0,4)$. $f$ is continuous.

(1) The function $f_1:[0,4) \rightarrow (0,4]$ defined by $f_1(x) = 4-x$ is continuous.

(2) The function $g_2: [0,2] \rightarrow [0,4]$ defined by $g_2(x) = x^2$ is continuous, increasing and surjective. Therefore its inverse $f_2^*: [0,4] \rightarrow [0,2]$ defined by $f_2^*(x) = \sqrt{x}$ is continuous. In particular, its restriction $f_2 =f_2^*|_{(0,4]}$ is continuous.

(3) The function $f_3:(0,4] \rightarrow \Bbb{R}$ defined by $f_3(x) = 1/x$ is continuous.

Pick any $x \in (0,\infty)$. We show that $f_3$ is continuous at $x$. Pick any $\epsilon >0$. Take $\delta = \min\{\epsilon x^2/2, x/2\}$.

Pick any $y \in (0,\infty)$ such that $|y - x| < \delta$. Then $y > x/2 > 0\implies 1/y<2/x$.

$$\bigg|\frac{1}{y}-\frac{1}{x} \bigg| = \bigg|\frac{x -y}{xy}\bigg| =\frac{|x -y|}{xy} < \frac{2\delta}{x^2} \leq \frac{2(\epsilon x^2/2)}{x^2} =\epsilon$$

Therefore $f_3$ is continuous at each $x \in (0,\infty)$.

And the function $f = f_3 \circ f_2 \circ f_1$ is continuous on $[0,4)$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .