6. Intuitive Limits and Continuity

To review: A function \(f(x)\) is continuous at \(x=a\) if: \[ \lim_{x\to a}f(x)=f(a) \] In more detail, the limit from the left, the limit from the right and the value at \(x=a\) all exist and are all equal. We frequently say “You can draw the graph and not pick up your pencil as you pass through \(x=a\).” More generally:

d. Continuity on an Interval

We say \(f(x)\) is continuous on the interval \([a,b]\) if it is continuous at \(x=c\) for all \(c\in[a,b]\). In formulas: \[ \lim_{x\to c}f(x)=f(c) \qquad \text{for all} \quad c\in[a,b] \] We frequently say “You can draw the graph without picking up your pencil anywhere in the interval \([a,b]\).”

The Limit Laws (discussed intuitively in the next chapter on Computing Limits and precisely in the subsequent chapter on Precise Limits) imply that all functions constructed using addition, subtraction, multiplication, division, exponentiation and logarithm are continuous provided all denominators are non-zero and all exponentiations and logarithms are defined. Consequently, all polynomials and rational functions (ratios of polynomials) are continuous on their domains. Further, all trig and inverse trig functions are continuous on their domains and all compositions of continuous functions are continuous. We will make this rigorous in the chapter on Precise Limits and Continuity.

Determine all points where the function \(g(x)=\dfrac{1}{\sqrt{4-x^2}}\) is continuous.

The function is continuous wherever the square root is defined and the denominator is non-zero. The square root is defined when \(4-x^2 \ge 0\) or when \(x^2 \le 4\) or \(|x| \le 2\). However, the denominator is zero when \(4-x^2=0\) or \(x=\pm2\). So \(g(x)\) is continuous on the open interval \(|x| \lt 2\) or \((-2,2)\) or \(-2 \lt x \lt 2\).