Prove $\lim\limits_{x\to -\infty} \frac{\cos(\tan x)}{x+1} = 0$ using definition

Question

Problem

Prove $\lim\limits_{x\to -\infty} \frac{\cos(\tan x)}{x+1} = 0$ using definition

proof

$\forall \varepsilon>0$ take $M=-\frac{1}{\varepsilon}-1$ if $x<M<0$, implies

$$x<-\frac{1}{\varepsilon}-1<0$$ it follows that $$-\varepsilon <\frac{1}{x+1}<0<\varepsilon \iff \bigg|\frac{1}{x+1}\bigg|<\varepsilon$$

since $|\cos(\tan x)|\leq 1$ it follows $$f(x)=\bigg|\frac{\cos(\tan x)}{x+1}\bigg|<\varepsilon$$ so $$\lim\limits_{x\to -\infty} \frac{\cos(\tan x)}{x+1} = 0$$

Please verify the proof above.

Answer 1

Your proof is fine, albeit only on $\{x<-1|\tan x\in\Bbb R\}$, a space where the limit makes sense. Alternatively (again subject to the same constraint @TheSilverDoe notes), squeeze between $\frac{\pm1}{x+1}$, each of limit $0$.

Answer 2

I make an answer because this will be too long for a comment.

Your proof is actually almost correct, except that you did not take into account the fact that your function is never defined on an interval of the form $(-\infty,a)$, with $a \in \mathbb{R}$. Indeed the function $$f(x)= \frac{\cos(\tan(x))}{x+1}$$ is not-defined in every $\frac{\pi}{2}-k\pi$ where $k\in \mathbb{Z}$ (because of the $\tan$), so there is no interval $(-\infty,a)$ on which $f$ is everywhere defined. So when you say "for every $x < M$, etc.", this is not really correct, because $x<M$ can be a point where the function is not-defined and in that case, what you write then has no sense.

What you must have in mind is that the definition of the limit takes this issue into account. For a function $g$ defined on a domain $\mathcal{D}$, the definition of $\lim_{x \rightarrow -\infty} g(x)=L$ is

$$\forall \varepsilon > 0, \exists M\in \mathbb{R}, \forall x \in (-\infty,M)\cap \mathcal{D}, |f(x)-L|< \varepsilon$$

The important point is "$\forall x \in (-\infty,M)\cap \mathcal{D}$", which is not just "$\forall x \in (-\infty,M)$". You only have to consider the $x$'s which belong to the domain of $g$. With this small precision, you can make an absolutely correct proof of your limit.

I just also want to add an often forgotten point about this definition of the limit in $\pm \infty$ : this definition has sense only if $\mathcal{D}$ is not bounded below (which is the case in your question, so no problem here). Consider for example the function $g : [0,1] \rightarrow \mathbb{R}$ defined by $g(x)=x^2$ on the domain $\mathcal{D}=[0,1]$. Then let $\varepsilon > 0$ and $M = -1$. You have $(-\infty,M) \cap \mathcal{D}= \emptyset$, so it is true that $\forall x \in (-\infty,M) \cap \mathcal{D}$, one has $|g(x)-42|<\varepsilon$, and also $|g(x)-127|<\varepsilon$ ! Therefore, just with the definition, you get that $$\lim_{x \rightarrow - \infty} g(x) = 42 \quad \quad \text{and} \quad \quad \lim_{x \rightarrow - \infty} g(x) = 127$$

ie $g$ has two distinct limits, which is a thing we don't want (an important property is that the limit, when it exists, must be unique). This happens here because $g$ is only defined on $[0,1]$ which is bounded below.