# Beginners calculus type of question (...I think)

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• Oct 28th 2010, 12:29 AM
jayshizwiz
Beginners calculus type of question (...I think)
Hi. I answered this question but I have no idea if I did it correctly...

Question: Find a $\delta > 0$ that fulfills: if $|x+1| < \delta$, then $\left| \frac{x}{2x+3} + 1 \right| < \frac{1}{100}$

Answer: $\left| \frac{x}{2x+3} + 1 \right|$ = $\left| \frac{3x+3}{2x+3} \right| = \frac{3|x+1|}{|2x+3|}$

That part was easy, now is the part I don't know if I did logically/correctly...

The function is not defined at $x = -\frac{3}{2}$ so we can require $0 \leq \delta \leq \frac{1}{5}$ (so the x can stay away from $-\frac{3}{2}$)

so that $|x+1| < \frac{1}{5}$ $\rightarrow$ $- \frac{1}{5} \leq x+1 < \frac{1}{5}$ $\rightarrow$ $0 < \frac{3}{5}< 2x+3 < \frac{7}{5}$ $\rightarrow$ $|2x+3|>\frac{3}{5}$

So now we get

$\frac{3|x+1|}{|2x+3|} \leq \frac{3|x+1|}{\frac{3}{5}} = 5|x+1|$

and since $|x+1| < \delta$, $5|x+1|<5\delta$

Therefore if we require $5\delta < \frac{1}{100}$, we will for sure get for $|x+1|< \delta$

$5|x+1| < 5\delta < \frac{1}{100}$

and so $0< \delta < \frac{1}{500}$

Thanks!

**edit thanks for your advice!
• Oct 28th 2010, 02:42 AM
Ackbeet
You were doing great until the very end. This line:
5|x+1| < \delta < \frac{1}{100}, should be
5|x+1| < 5\delta < \frac{1}{100}. You corrected it in the next line, though.
Also, you found that delta < 1/500. Great! Stop right there. You're done. delta < 1/5 does NOT satisfy the requirement.
• Oct 28th 2010, 04:40 AM
jayshizwiz
Quote:

Originally Posted by Ackbeet
You were doing great until the very end. This line:
5|x+1| < \delta < \frac{1}{100}, should be
5|x+1| < 5\delta < \frac{1}{100}. You corrected it in the next line, though.

Thanks! That was a typo. I'll edit it.

Quote:

Also, you found that delta < 1/500. Great! Stop right there. You're done. delta < 1/5 does NOT satisfy the requirement
Thanks again. I just got confused since my initial statement was (1) $0 \leq \delta \leq \frac{1}{5}$ but I guess since my final result was (2) $\delta < \frac{1}{500}$, as long as I choose a number smaller than (2), it also satisfies (1)....(as long as it's bigger than 0...)
• Oct 28th 2010, 04:44 AM
Ackbeet
Looks like you're good to go, then. Yeah, you have to take the more restricted condition in cases like this.