Proving g(x) is continuous over interval (-∞,-2)

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Homework Statement
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Relevant Equations
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For number 18,
1676597607436.png

The solution is,
1676597636101.png

However, should they not write "For ## -∞ < a < -2##" since ##a ≠ -∞## (infinity is not a number)?

Many thanks!
 
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##a## is a number, so can't be negative infinity.

In particular, ##a\in [-\infty,-2)## isn't really a meaningful thing outside of people abusing notation.
 
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Office_Shredder said:
##a## is a number, so can't be negative infinity.

In particular, ##a\in [-\infty,-2)## isn't really a meaningful thing outside of people abusing notation.
Thank you for your reply @Office_Shredder !

So a better notation than ## a < -2 ## is ## -∞ < a < -2##, correct?

Many thanks!
 
Callumnc1 said:
Thank you for your reply @Office_Shredder !

So a better notation than ## a < -2 ## is ## -∞ < a < -2##, correct?

Many thanks!
No, the ##-\infty<a## notation conveys no additional information. ##a## and ##a>-\infty## are the same thing. You can include the negative infinity, but in no sense is it better here.
 
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Callumnc1 said:
However, should they not write "For −∞<a<−2"
##-\infty < a < -2## and ##a \in (-\infty, -2)## are two different notations that say exactly the same thing. In most of the books I've seen, interval notation, as in the 2nd example above, uses a parenthesis to indicate an endpoint that isn't included, and a bracket to indicate that an endpoint is included. I've seen other notations used, but these seem to be a lot rarer.

Also, I don't think any textbook would include "For" in a compound inequality. ##-\infty < a < -2## says everything that needs to be said.
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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