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

Question

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. $$|f(x) -L| = \cdots = cx > cN=c\frac{\epsilon}{c} = \epsilon$$

Answer 1

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.

Answer 2

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 $x \ge 35$.

Answer 3

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}

Answer 4

Let $\epsilon >0$. We are searching for a $c>0$, such that $x\geq c\Rightarrow |\frac {x+1}{x+5} -1|<\epsilon \iff |\frac {-4}{x+5}|<\epsilon \iff \frac {4}{\epsilon}<x+5 \iff \frac {4}{\epsilon} -5<x$. 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 $|\frac {x+1}{x+5} -1|<\epsilon \iff |\frac {-4}{x+5}|<\epsilon \iff \frac {4}{\epsilon}<x+5 \iff \frac {4}{\epsilon} -5<x$ 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.