3
$\begingroup$

Use the $\varepsilon$ - $\delta$ definition of limit to prove that $\displaystyle\lim_{x\to\,-1}\frac{x}{2x+1}=1$.

My working:

$\left|\frac{x}{2x+1}-1\right|=\left|\frac{-x-1}{2x+1}\right|=\frac{1}{\left|2x+1\right|}\cdot \left|x+1\right|$

First restrict $x$ to $0<\left|x+1\right|<\frac{1}{4}$ $\Rightarrow$ initial choice of $\delta=\frac{1}{4}$

$\left| 2x+1 \right|$ = $\left|2(x+1)-1\right|$

$\le$ $\left|2(x+1)\right|+\left|-1\right|$ = $2\left|x+1\right|+1$ $> 1$

Thus if $\left|x+1\right|<\frac{1}{4}$ , then

$\left|\frac{x}{2x+1}-1\right|=\frac{1}{\left|2x+1\right|}.\left|x+1\right|$ $<1.\left|x+1\right|$.

Therefore, $\delta = \min\{\frac{1}{4},\varepsilon\}$

$0<\left|x+1\right|<\delta$ $\Rightarrow$ $\left|\frac{x}{2x+1}-1\right| < 1\cdot\left|x+1\right| < 1\cdot\varepsilon = \varepsilon$

Thus, the limit is 1.

$\endgroup$
  • $\begingroup$ you say above you have numbers $a\le b>c$, and then conclude that $a>c$. This is not a correct conclusion, for instance, $0\le2>1$, but $0\not>1$ $\endgroup$ – Brady Trainor Mar 23 '13 at 17:39
  • $\begingroup$ You need to write your expression in terms of $x + 1$ so the last steps get easier. $\endgroup$ – vonbrand Mar 23 '13 at 17:48
  • $\begingroup$ You should have written $|2x+1|\le2|x+1|+1<2\cdot{1\over 4}+1$ $\endgroup$ – Brady Trainor Mar 23 '13 at 17:57
  • $\begingroup$ but that's not right either, you need a lower bound on $|2x+1|$, because it's in the denominator $\endgroup$ – Brady Trainor Mar 23 '13 at 18:03
4
$\begingroup$

To "discover" the proof, we typically work backwards. We "assume" ${|x+1|\over|2x+1|}<\epsilon$, and find what $\delta$ should be. Then we will have to rewrite the proof.

Let's think about the numerator and denominator separately. We want $|x+1|$ small, and intuitively, we want $|2x+1|$ large. So let's keep $2x+1$ away from $0$ first. I think that's the $\delta<{1\over4}$ part. Then $$-{1\over 4}<x-(-1)<{1\over4},$$ so $-{5\over 4}<x<-{3\over4}$, so $-{5\over 2}<2x<-{3\over2}$ so $-{3\over 2}<2x+1<-{1\over2}$. So we use the fact that $|2x+1|>{1\over2}$. Thus, $${|x+1|\over|2x+1|}<2|x+1|<\epsilon.$$ Then we notice $$\delta<\min\bigg\{{\epsilon\over2},{1\over4}\bigg\}$$

Should do the trick. At this point, we've just "discovered" the proof, and $\delta(\epsilon)$, so now we proceed to write the logic out concisely. Fix $\epsilon$, blah blah blah

$\endgroup$
1
$\begingroup$

You shouldn't change direction of your inequalities in a chain--for example, $$|2(x+1)-1|\le 2|x+1|+1>1$$ doesn't allow you to conclude that $|2(x+1)-1|>1,$ as transitivity breaks down when you switch directions.

Instead, we can use triangle inequality (why?) to say $$|2(x+1)-1|\ge1-2|x+1|,$$ so whenever $|x+1|<\frac14,$ we will have $$|2(x+1)-1|\ge1-2|x+1|>\frac12,$$ and so $$\left|\frac{x}{2x+1}-1\right|=\frac1{|2(x+1)-1|}|x+1|<2|x+1|.$$ A good choice would then be $$\delta=\min\left\{\frac14,\frac{\epsilon}2\right\}.$$

$\endgroup$
0
$\begingroup$

Hint: Note that \begin{align} \left| \frac{x}{2\cdot x +1} - 1 \right| < \epsilon \Longleftrightarrow & 1-\epsilon< \frac{x}{2\cdot x +1}< 1+\epsilon \\ \Longleftrightarrow & \left\{ \begin{array} (2x+1)(1-\epsilon)<x \\ \\ x< (2x+1)(1+\epsilon) \end{array} \right. \end{align} After algebric manipulations of two last inequalities you get $$ \left| \frac{x}{2\cdot x +1} - 1 \right| < \epsilon \Longleftrightarrow \left\{ \begin{array} \;x-1<\frac{1+\epsilon}{1+2\epsilon}\\ \\ x-1< \frac{1-\epsilon}{1-2\epsilon} \end{array} \quad \mbox{ for } \epsilon <\frac{1}{2} \right. $$ This suggests $\delta=\min\{\frac{1-\epsilon}{1-2\epsilon},\frac{1-\epsilon}{1-2\epsilon}\}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.