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Example 6 and Example 7.2

Are example 6 and example 7.2 correct? Shouldn't be example 6 be (-infinity,1] U [1,+infinity), and should be example 7.2 be x≥4?

Janjan
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  • Welcome to MSE! Please use MathJax. Please do not use pictures for critical portions of your post. Pictures may not be legible, cannot be searched and are not view-able to some, such as those who use screen readers. – GNUSupporter 8964民主女神 地下教會 Dec 27 '17 at 17:28
  • Uh. No.... If it is not defined for any $x > 1$ how can it be continuous at $x = 1$. – fleablood Dec 27 '17 at 17:29
  • http://www.math.psu.edu/tseng/class/Math140A/Notes-Continuity.pdf –  Dec 27 '17 at 17:31
  • show your efforts... – Learning Dec 27 '17 at 17:31
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    In 7.2, $x=4$ is the point where the piecewise function breaks into pieces, so we must be careful there. The analysis done in 7.2 only shows that $f(x)$ is continuous for $x > 4$. The function is indeed continuous at $x=4$ also, but you have to show that by showing that $\displaystyle\lim_{x\to4} f(x) = f(4)$. To do that you need to evaluate both one-sided limits. Also, what exactly is the question for #7? It just says "consider the function..." and then goes into the solution. Solution to what? Largest interval of continuity again? –  Dec 27 '17 at 17:33
  • 7.1 is a temporary step to show that f is continuous for x > 4. And 7.2 is a temporary step to show that f is continuous for x < 4. I assume 7.3 will be a step to show that f is continuous at x = 4. 7.2 does not tell us anything about the continuity at x = 4. – fleablood Dec 27 '17 at 17:47

2 Answers2

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Your text is using the definition of continuous at $a$ to mean

$\lim_{x\to a} f(x) = f(a)$.

for $\lim_{x\to a} f(x) = K$ to be true two things must happen. $\lim{x \to a; x > a} f(x) = K$ and $\lim{x\to a; x < a} f(x) = K$.

In 6)

$\lim_{x\to -1; x > -1} f(x)$ does not exist. so $\lim_{x \to -1} f(x) \ne f(-1)$ so $f$ is not continuous at $x = -1$. Likewise $\lim_{x\to 1; x < 1}f(x)$ does not exist so $\lim_{x\to 1}f(x) \ne f(1)$.

As for 7)

$\lim_{x\to 4; x > 4} = \lim_{x\to 4} x-1 = 3$ whil $\lim_{x\to 4; x < 4} = \sqrt{x} + 1 = 3$ and $f(4) = \sqrt{4} + 1 = 3$. So $\lim_{x\to 4} f(x) = f(4)$ so $f$ is continuous at $4$.

.... and I assume that is step 3) on the next page.

7.1 is a temporary step to show that $f$ is continuous for $x > 4$. And 7.2 is a temporary step to show that f is continuous for $x < 4$. I assume 7.3 will be a step to show that f is continuous at$ x = 4$. 7.2 does not tell us anything about the continuity at $x = 4$.

fleablood
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You are correct. x=1 and x=-1 should be included in the solution for example 6 and x=4 should be included in the solution for 7.2. Functions are continuous at those points.

Mohammad Riazi-Kermani
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  • In 7, f is continuous at x=4 and I assume the text will show that in 7.3. But in 6. f is not continuous at x =1 or x= -1 because the limits are only one sides. For any $\delta > 0$ then $-1 < x < -1 + \delta \implies |\sqrt{x^2-1} - f(-1)|$ does not exist. Hence f is not continuous at $-1$. – fleablood Dec 27 '17 at 17:56