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Let $f$ be the function, of domain $\mathbb{R}$, represented on the picture and let $(u_n)$ be a sequence.

It is known that $\lim f(u_n) = 2$

enter image description here

Which of the following expressions could be the rule of $(u_n)$?

a) $2-\frac{1}{n}$

b) $\frac{1}{n}-2$

c) $n-2$

d) $2-n$

Here's what I tried:

  • If $\lim f(u_n) = 2$, then when $f$ tends to 2 that means it is tending to $u_n$ (I think)
  • $f$ tends to 2 when $x \rightarrow -\infty$ and $x = 2$
  • That means that $u_n$ has in its output $-\infty$ and $2$
  • This eliminates options a) and b)

I don't know what to do next though. Am I doing this correctly so far?

How do I solve this? My book says the solution is d)

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    Since $\lim_{x\to -\infty }f(x)=2$ and that $\lim_{n\to \infty }2-n$, then of course that $\lim_{n\to \infty }f(2-n)=2$. By the way $\lim_{n\to \infty }f(n-2)=0$ since $\lim_{x\to \infty }f(x)=0$. Therefore d) is the only possibility. – 2017-02-11

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Heine's definition of limit says $$\lim_{x\rightarrow a}f(x)=L \Leftrightarrow \forall x_n\rightarrow a, f(x_n)\rightarrow L.$$

Now try the options (I'll do some examples):

(a) $u_n\rightarrow 2^-$ and $\lim\limits_{x\rightarrow 2^-}f(x)=-\infty$ so $f(u_n)\rightarrow -\infty$ so this one is false.

(d) $u_n\rightarrow -\infty$ and $\lim\limits_{x\rightarrow -\infty}f(x)=2$ so $f(u_n)\rightarrow 2$. This one works.

Can you now show that (b), (c) are also false?

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    I will try but I have one question first; where you wrote (a) $u_n\rightarrow 2^-$ don't you mean $u_n\rightarrow 2$ instead? – 2017-02-11
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    I mean that $u_n$ approaches $2$ from the left, i.e. it approaches $2$ by values which are lesser than $2$. – 2017-02-11
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    I have been trying to solve this like you said using several limits (mostly $n \rightarrow +\infty$, $n \rightarrow -\infty$ and $n \rightarrow 2$) but for (d) I get $\lim_{n \rightarrow - \infty}u_n = + \infty$ and so $u_n \rightarrow +\infty$ and $\lim\limits_{x\rightarrow +\infty}f(x)=0$ which means $f(u_n)\rightarrow 0$. What did I do wrong? – 2017-02-11
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    Also, can $a$ be any value of the codomain or any limit of $u_n$? – 2017-02-11
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    When we talk about sequences it only makes sense to talk about its limit when $n\rightarrow +\infty$, because $n\in\mathbb{N}$. – 2017-02-11
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    So when I say $u_n\rightarrow\mbox{ something}$ I mean the limit when $n\rightarrow +\infty$ – 2017-02-11
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    Ok I understand now. One last thing, why does $\lim_{n \rightarrow +\infty} u_n = 2^-$ for (a)? Why not $\lim_{n \rightarrow +\infty} u_n = 2^+$? Or simply $\lim_{n \rightarrow +\infty} u_n = 2$ ? – 2017-02-11
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    The limit is $2$, that notation just means that the sequence is approaching $2$ by values which are lesser than $2$ $(1, 1.5, ...)$. This is useful because you have to compute the limit of $f$ when $x$ goes to $2$ but that one does not exist; since you know $u_n$ goes to $2$ "from the left", what you need is the left-handside limit of $f$ at $2$. – 2017-02-11
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    For example, $u_n=2+1/n$ would have limit $2^+$ (again, just notation!), so $$\lim f(u_n)=\lim\limits_{x\rightarrow 2^+}f(x)=2.$$ – 2017-02-11
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    Yes, I understand the notation now, what I was asking was if $\lim_{n\rightarrow 2^+}2-\frac{1}{n}$ exists or if it is indeterminate. As for proving that (b) and (c) are false, I think I got it: (b) $u_n\rightarrow -2^-$ and$ \lim\limits_{x\rightarrow -\infty}f(x) = a, a \in ]1;2[$ as in a is some value between 1 and two (if this notation is incorrect please let me know) and (c) $u_n\rightarrow + \infty$ and $\lim\limits_{x\rightarrow +\infty}f(x) = 0$ – 2017-02-12