2
$\begingroup$

I need to find the following limit and prove using the definition of limits.

$$\lim_{x\to1} {x \over x+1} = \frac 1 2$$.

Following the definition: $$\forall \epsilon \exists \delta : \lvert x - c \rvert < \delta \Rightarrow \lvert F(x) - L \rvert < \epsilon$$

$$\left\lvert \frac{x}{x+1} - \frac{1}{2} \right\rvert < \epsilon = \left\lvert 2x-x-1 \over 2x+2 \right\rvert = \left\lvert x-1 \over 2x+2 \right\rvert < \epsilon$$

I have trouble around here. I don't know how to reach $\left\lvert x - c \right\rvert < \delta$

I tried:

$$ \frac{x-1}{2x+2} < \frac{x}{2x} = \frac{1}{2} < \epsilon$$

But something about that doesn't seem right to me. Can I get any hints?

$\endgroup$
2
  • $\begingroup$ You've missed all your dollar signs. I'm not able to edit it. $\endgroup$ Nov 21 '14 at 9:48
  • $\begingroup$ This is a similar question $\endgroup$
    – John
    Nov 21 '14 at 9:57
1
$\begingroup$

Hint:

Take $\delta=min(\epsilon,1)$, then $\forall |x-1|<\delta\le 1$, $|x+1|>1$, hence ...

The idea is to select $\delta$ such that $x$ is closer to $1$ than to $-1$ (the denominator is $2|x+1|$), so that you have a lower bound for the denominator.

$\endgroup$
2
  • $\begingroup$ Sorry, but you could you elaborate? I'm having trouble wrapping my head around it... Edit: Nevermind! I got it! Thanks! $\endgroup$
    – user194351
    Nov 21 '14 at 10:10
  • $\begingroup$ NB I fixed a minor grammar issue / typo. $\endgroup$
    – AlexR
    Nov 21 '14 at 10:11
0
$\begingroup$

Showing the versatility of this problem.

$$\begin{align}|x - 1| < \delta \leq \frac{1}{2} &\Leftrightarrow -\frac{1}{2} < x - 1 < \frac{1}{2} \Leftrightarrow \frac{1}{2} < x < \frac{3}{2} \\ &\Leftrightarrow \frac{3}{2} < x + 1 < \frac{5}{2} \Rightarrow \frac{2}{5} < \frac{1}{x+1} < \frac{2}{3} \end{align}$$

In particular $\Big|\frac{1}{x+1}\Big| < \frac{2}{3}$.

Then as you have already reached

$$\left\lvert \frac{x}{x+1} - \frac{1}{2} \right\rvert = \left\lvert 2x-x-1 \over 2x+2 \right\rvert = \left\lvert x-1 \over 2x+2 \right\rvert = \frac{1}{2}\left\lvert x-1 \over x+1 \right\rvert < \frac{\delta}{3}$$

Now take $\delta = \min \lbrace \epsilon , \frac{1}{2}\rbrace$. And you will have for every $\epsilon > 0$ given, there exits $\delta = \min \lbrace \epsilon, \frac{1}{2}\rbrace$ such that

$$|x - 1| < \delta \Rightarrow \Bigg|\frac{x}{x + 1} - \frac{1}{2}\Bigg| < \epsilon$$

$\endgroup$

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.