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

Question

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.

Answer 1

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 $$\delta=\min\left\{\frac14,\frac{\epsilon}2\right\}.$$

Answer 2

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

Answer 3

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 $\delta=\min\{\frac{1-\epsilon}{1-2\epsilon},\frac{1-\epsilon}{1-2\epsilon}\}$.