By using the definition of limit only, prove that $\lim_{x\rightarrow 0} \frac1{3x+1} = 1$

By using the definition of a limit only, prove that

$$\lim_{x\rightarrow 0} \dfrac{1}{3x+1} = 1$$

We need to find $$0<\left|x\right|<\delta\quad\implies\quad\left|\dfrac{1}{3x+1}-1\right|<\epsilon.$$ I have simplified $$\left|\dfrac{1}{3x+1}-1\right|$$ down to $$\left|\dfrac{-3x}{3x+1}\right|$$

Then since $${x\rightarrow 0}$$ we can assume $$-1 then $$-2<3x+1<4$$ which implies $$\left|\dfrac{1}{3x+1}-1\right|=\left|\dfrac{-3x}{3x+1}\right|<\left|\dfrac{-3x}{4}\right|<\left|\dfrac{-3\delta}{4}\right|=\epsilon$$ No sure if the solution is correct

• The step $$\left|\dfrac{-3x}{3x+1}\right|<\left|\dfrac{-3x}{4}\right|$$ is not correct. You said $3x+1<4$. If you are dividing $|-3x|$ by a bigger number, you get a smaller result.
– Ovi
Aug 14, 2017 at 21:59

If we focus on $-1<x<1$ and end up with $-2<3x+1 < 4$, $\frac{1}{3x+1}$is unbounded.

Rather than focusing on $-1 < x < 1$, focus on a smaller interval, for example $|x| < \frac14$. Hence $\delta < \frac14$.

if $-\frac14 < x < \frac14$, $$-\frac34+1 < 3x+1< \frac34+1$$.

$$\frac14 < 3x+1< \frac74$$

$$\left|\frac{1}{3x+4} \right| < 4$$

Hence $$12\delta < \epsilon$$

Note that

$$\left|\frac{1}{1+3x}-1\right|=\left|\frac{3x}{1+3x}\right| \tag 1$$

Now, we restrict $x$ such that $x\in [-1/4,1/4]$. And with this restriction, it is easy to see that $1/4/ \le 1+3x$. Using this in $(1)$ reveals that

$$\left|\frac{1}{1+3x}-1\right|\le 12|x|\tag 2$$

Finally, given any $\epsilon>0$,

$$\left|\frac{1}{1+3x}-1\right|<\epsilon$$

whenever $|x|<\delta =\min\left(\frac14,\frac{\epsilon}{12}\right)$.

• Please let me know how I can improve my answer. I really want to give you the best answer I can. And once you've accrues enough reputation points, you can up vote answers as you see fit. Aug 18, 2017 at 16:00

No. If we assume $-1 < x < 1$, then $-2 < 3x + 1 < 4$. Since $3x+1=0$ is within those limits, then $\left|\dfrac {1}{3x+1}\right|$ will be unbounded.

If you want to put an upper bound on $\left|\dfrac {1}{3x+1}\right|$, then you are going to need something like $3x + 1 > \frac 13$, which is equivalent to $x > -\frac 29$.

So, $|x| < \frac 29 \implies 3x+1 > \frac 13 \implies \left|\dfrac {1}{3x+1}\right|<3$.

So, then, $\left|\dfrac{1}{3x+1}-1\right|=\left|\dfrac{-3x}{3x+1}\right| < |9x|$

To make that less than $\epsilon$, we need to have $|x|<\frac 19 \epsilon$.

So we need to have $\delta < \frac 19\epsilon$ and $\delta < \frac 29$.

Hence we define $\delta = \min(\frac 19\epsilon, \frac 29)$