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- Thread starter adelin
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- #1

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- #2

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If so, can you start by explaining what you think they're doing? And can you explain what you don't get??

- #3

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If so, can you start by explaining what you think they're doing? And can you explain what you don't get??

This is another proof

- #4

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This is another proof

It is very similar. So please, tell us what you think first.

- #5

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It is very similar. So please, tell us what you think first.

in the next step they arrive to the same conclusion. If δ <1 then δ<10. ( I may be wrong)

The next step is what become problematic for me to understand.

- #6

Mark44

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In the second line, which is what I believe you're asking about, they make the assumption that ##\delta < 1##. Then if ##|x - 4 | < \delta < 1##, they can say that x will be between 3 and 5. Note that I'm ignoring the part where it says 0 < |x - 4|. All this does is eliminate the possibility of x being equal to 4.

Since 3 < x < 5, the largest that |x + 5| can be is 10. From this, they can write

## |x + 5||x - 4| < 10|x - 4|##

If we take ##\delta = \epsilon/10##, then when ##|x - 4| < \delta##, it will follow that

##|x + 5||x - 4| < 10|x - 4| < 10 * \delta < 10 * \epsilon/10 = \epsilon ##

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