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

1. ## 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!

2. 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.

3. 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.

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...)

4. Looks like you're good to go, then. Yeah, you have to take the more restricted condition in cases like this.