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Could anyone tell me how $|f'(x)- L| < |L|/2$ give us $|f'(x)| \gt |L|/2$ ? and $L \ne 0$.

Emptymind
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    It is not true. You haven't given enough context. $|0 - 1| > 1/2$, but $|0| < 1/2$. – Aryabhata Aug 02 '17 at 07:11
  • Thank u I felt it is not true as u said, but someone wrote it in an answer for another question and insist that it is right ..... so could u tell me the truth please? @Aryabhata – Emptymind Aug 02 '17 at 07:14
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    Give a link to that answer please. Context changes everything. – Aryabhata Aug 02 '17 at 07:16
  • @ Aryabhata https://math.stackexchange.com/questions/2378668/limits-at-infinity-of-a-function-and-its-derivative/2378679?noredirect=1#comment4905532_2378679 the accepted answer. – Emptymind Aug 02 '17 at 07:31
  • @Aryabhata the link is in the above comment. – Emptymind Aug 02 '17 at 07:34
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    You have interchanged the $\lt$ with $\gt$ in $|f'(x) - L| \lt |L|/2$. Basically, the distance between $f'(x)$ and $L$ is not more than $|L|/2$. If you take the interval $(L - |L|/2, L+|L|/2)$, $f'(x)$ must lie in it. – Aryabhata Aug 02 '17 at 07:37
  • yeah u are right .... I am sorry ..... I will correct it...... @Aryabhata – Emptymind Aug 02 '17 at 07:39
  • but the last inequality in the question did not say what u said in your last comment @Aryabhata – Emptymind Aug 02 '17 at 07:45
  • Yes, that last comment gives you some insight when L is positive and so $|L| = L$. Similarly consider the case then $|L| = -L$ – Aryabhata Aug 02 '17 at 07:53

1 Answers1

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A simple proof is to use

$$|a - b| \ge |a| - |b|$$

$$|L|/2 \gt |f'(x) - L| = |L - f'(x)| \ge |L| - |f'(x)|$$

Thus

$$|f'(x)| + |L|/2 \gt |L| \implies |f'(x) \gt |L| - |L|/2 = |L|/2$$

Aryabhata
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