# Delta Epsilon Proof $\lim_{x\to \infty} \frac{x+1}{x+5} =1$

I am trying to prove the following limit using the delta epsilon definition, $$\lim_{x\to \infty} \frac{x+1}{x+5} =1$$ So I want to prove that $$\forall N>0, \exists \epsilon >0| x >N \rightarrow \frac{x+1}{x+5}-1 < \epsilon$$

I assume I don't need absolute value for $$f(x)-L$$ since $$x \to \infty$$.

Now I can do some scratchwork as follows, $$\frac{x+1}{x-5} -1 < \epsilon \rightarrow x > \frac{-4-5\epsilon}{\epsilon}$$ Now I can begin the proof as follows,

Given $$\epsilon >0$$, choose $$N= \frac{-4-5\epsilon}{\epsilon}$$

For $$x>N$$ $$x > \frac{-4-5\epsilon}{\epsilon}\rightarrow ?$$

But I'm not sure where to go from here, even after rearranging to isolate for x. This seems complicated because I have to work my way backwards to the original fraction $$\frac{x+1}{x+5} -5 < \epsilon$$. Is there anyway I could manipulate this sort of question to To start with the original fraction/limit statement and work towards isolating x and then substituting N? E.g. $$|f(x) -L| = \cdots = cx > cN=c\frac{\epsilon}{c} = \epsilon$$

• Try to use some inequalities. Dec 5, 2018 at 4:50

You have a wrong definition of limit. It's not $$\forall N>0, \exists \epsilon >0\, | x >N \rightarrow \frac{x+1}{x+5}-1 < \epsilon$$ It's $$\forall \epsilon>0, \exists N >0\, | x >N \rightarrow \left|\frac{x+1}{x+5}-1\right| < \epsilon$$ By the way, you can't just ditch the absolute value.

• "I assume I don't need absolute value for f(x)−L since x→∞.", is that not true? Dec 5, 2018 at 5:06
• Of course not. $|f(x)-L|<\epsilon$ means $f(x)$ is near $L$. $f(x) - L < \epsilon$ could happen if $f(x)-L=-1000000$ Dec 5, 2018 at 5:09

Given $$\Delta y$$, find $$x_0$$ s.t. $$\forall x > x_0$$ :

$$\left|\frac{x + 1}{x+5} - 1\right| \le \Delta y$$

Hopefully they are still teaching kids how to divide polynomials in school:

$$\left|1 + \frac{-4}{x+5} - 1\right| \le \Delta y$$ $$\left|\frac{-4}{x+5}\right| \le \Delta y$$

Since we only care about sufficiently large $$x$$, we can assume $$x+5$$ is positive, so:

$$\frac{\color{red}+4}{x+5} \le \Delta y$$

(and you can see now that you can't just "drop" the absolute value without consideration)

$$\frac{4}{\Delta y} - 5 \le x$$

So if I say "$$\dfrac{x+1}{x+5}$$ within 0.1 of 1" you can say $$\dfrac{4}{0.1} - 5 = 35$$, so whenever $$x \ge 35$$.

Hint: For sufficiently "big" $$x$$ such that $$5 < \frac{1}{2}|x|$$, i.e., $$|x| > 10$$, it follows that: \begin{align} \left|\frac{x + 1}{x + 5} - 1\right| = \left|\frac{-4}{x + 5}\right| = \frac{4}{|x + 5|} \leq \frac{4}{|x| - 5} < \frac{8}{|x|}. \end{align}

• I see how I could work with this using $\frac{1}{x} <\frac{1}{N}$, however what did you do in this step $\frac{4}{|x+5|} \leq \frac{4}{|x| -5}$? Dec 5, 2018 at 4:58
• @Danielle Triangle inequality: $|x + 5| \geq ||x| - 5| = |x| - 5$. By the way, I assume your $\infty$ includes both $+\infty$ and $-\infty$, so the absolute value sign is kept throughout. Dec 5, 2018 at 5:08

Let $$\epsilon >0$$. We are searching for a $$c>0$$, such that $$x\geq c\Rightarrow |\frac {x+1}{x+5} -1|<\epsilon \iff |\frac {-4}{x+5}|<\epsilon \iff \frac {4}{\epsilon}. Setting $$c:=max\{\frac 45,\frac {4}{\epsilon} -5\}$$ we have completed the proof.

Explanation. We firstly suppose an arbitrary $$\epsilon>0$$. We need to find a $$c>0$$, such that $$x\geq c\Rightarrow |\frac {x+1}{x+5} -1|<\epsilon$$ (this is just the definition of limit). The written equivalences $$|\frac {x+1}{x+5} -1|<\epsilon \iff |\frac {-4}{x+5}|<\epsilon \iff \frac {4}{\epsilon} mean that, if I start from the last inequality, I can get the first. So it could be beneficial to get inspiration from last inequality to choose a $$c>0$$: the number $$\frac {4}{\epsilon}-5$$ is $$0$$, when $$\epsilon=\frac 45$$ and it is positive, when $$\epsilon<\frac 45$$. Consequently I can choose the desirable $$c>0$$ as the maximum of $$\frac 45, \frac {4}{\epsilon}-5$$, because, if $$\epsilon$$ is more than or equal to $$\frac 45$$, the number $$\frac {4}{\epsilon}-5$$ is non positive (so undesirable); if $$\epsilon$$ is less than $$\frac 45$$, the number $$\frac {4}{\epsilon}-5$$ is quite big for our choice of $$c>0$$ (we need $$x\to \infty$$). You can easily attest that

$$x\geq c=max\{\frac 45, \frac {4}{\epsilon}-5\}\Rightarrow x\geq \frac {4}{\epsilon}-5\iff |\frac {x+1}{x+5}-1|<\epsilon$$, q.e.d.

• Is it possible for you to split this is to more sentences, to make it easier to understand? Sep 30, 2021 at 10:06