6. Intuitive Limits and Continuity

To review: A function f(x)f(x) is continuous at x=ax=a if: limxaf(x)=f(a) \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=ax=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=ax=a.” More generally:

d. Continuity on an Interval

We say f(x)f(x) is continuous on the closed interval [a,b][a,b] if it is continuous at x=cx=c for all c(a,b)c\in(a,b), right continuous at x=ax=a and left continuous at x=bx=b. In formulas: limxcf(x)=f(c)for allc(a,b)limxa+f(x)=f(a)andlimxbf(x)=f(b)\begin{aligned} \lim_{x\to c}f(x)&=f(c) \quad \text{for all} \quad c\in(a,b) \\ \lim_{x\to a^+}f(x)&=f(a) \quad \text{and} \quad \lim_{x\to b^-}f(x)&=f(b) \end{aligned} We frequently say “You can draw the graph without picking up your pencil anywhere in the interval [a,b][a,b].”
Similar definitions hold for open or half open intervals.

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, root and logarithm are continuous provided all denominators are non-zero and all exponentiations, roots 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)=14x2g(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 4x204-x^2 \ge 0 or when x24x^2 \le 4 or x2|x| \le 2. However, the denominator is zero when 4x2=04-x^2=0 or x=±2x=\pm2. So g(x)g(x) is continuous on the open interval x<2|x| \lt 2 or (2,2)(-2,2) or 2<x<2-2 \lt x \lt 2.

Find all points where the function f(x)=1x24xf(x)=\dfrac{1}{\sqrt{x-2}}-\sqrt{4-x} is continuous.

Hint

The function is continuous when both square roots are defined and the first denominator is non-zero.

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Answer

f(x)\displaystyle f(x) is continuous on the half-open interval (2,4](2,4].

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Solution

To make the square roots well-defined, we need x20x-2 \ge 0 and 4x04-x \ge 0. This says x2x \ge 2 and 4x4 \ge x. In other words: 2x42 \le x \le 4. However, if x=2x=2 the denominator will be 00. So the function is continuous on the half-open interval (2,4](2,4].

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