Let ($u_n$) and ($v_n$) be the successions defined by:

$u_n = \frac{n+1}{n}$ and $v_n = \frac{n-1}{n}$

It is known that:

$\lim f(u_n) = 0$ and $\lim f(v_n)=2$

a. What can be concluded about $\lim_{x \rightarrow 1}f(x)$?

b. Can the function $f$ have any lateral limits at point 1? If yes, what do you think they are? Justify.

According to my book, the answers are:

a. It doesn't exist.

b. Yes, 0 and 2.

My book doesn't specify what limit $\lim$ is exactly, but I think it is $\lim_{x \rightarrow \infty}$

I know that both $u_n$ and $v_n$ tend to 1 because their horizontal asymptote is y=1.

My questions are:

  1. If both $u_n$ and $v_n$ tend to the same value then how is it possible that $\lim f(u_n)$ and $\lim f(v_n)$ are different values?
  2. Could you define $f(x)$ ?
  3. How do you conclude that $f(x)$ is undefined where $x=1$ ? Am I supposed to assume so just because $u_n$ and $v_n$ are? Is this problem missing information?
  4. What is exactly a lateral limit? Is it an asymptote? If yes, is it a vertical or horizontal one? (I know what a limit is, I'm just not sure of what a lateral one might be)
  5. Since my book failed to do so, could you justify b) ?


  • $\begingroup$ It's limit from different paths one from left i.e $1^-$ and another one from right $1^+$. $\endgroup$ – kingW3 Jan 20 '17 at 18:35
  • $\begingroup$ An example of the function is $$\frac{2}{1+2^{1/(x-1)}}$$ and for $x=1$ take $f(1)=c$ for any $c$.Also the function can be defined at $x=1$ but it isn't continuous at $x=1$ $\endgroup$ – kingW3 Jan 20 '17 at 20:04

to answer this question it is helpful to have a clear definition of the different types of limits involved (the limit of a sequence, the normal limit and the one-sided limits which is probably meant by 'lateral limits' in your book):

As stated for example here the left- and right-hand limits are defined as

  1. Right-hand limit: $\displaystyle \lim_{x \to a^+} f(x) = L$ if and only if

For every $\epsilon>0$ there exists a $\delta>0$ such that for all $x$ with $0 < x-a <\delta$: $|f(x)-L| < \epsilon$.

  1. Left-hand limit: $\displaystyle \lim_{x \to a^-} f(x) = L$ if and only if

For every $\epsilon>0$ there exists a $\delta>0$ such that for all $x$ with $0 < a-x <\delta$: $|f(x)-L| < \epsilon$.

  1. Limit $\displaystyle \lim_{x \to a} f(x) = L$ if and only if

For every $\epsilon>0$ there exists a $\delta>0$ such that for all $x$ with $0 < |a-x| <\delta$: $|f(x)-L| < \epsilon$.

  1. Limit of a sequence: Let $a_n\in\mathbb{R}$ for all $n\in\mathbb{N}$ be a sequence of real numbers. Then $\lim_{n\rightarrow\infty} a_n=L$ if and only if

For every $\epsilon>0$ there exists $N\in\mathbb{N}$ such that for all $n>N$: $|a_n-L| < \epsilon$.

Given this definition and by noting that $(|a-x| < \delta)\Leftrightarrow(0<a-x < \delta\lor0<x-a < \delta)$ it is an immediate consquence that $\lim_{x \to a} f(x) = L$ if and only if $\lim_{x \to a^{+}} f(x)=L$ and $\lim_{x \to a^{-}} f(x)=L$.

Now to answer your questions:

  1. This is possible because for all $n\in\mathbb{N}$ $u_n>1$ and $v_n<1$ so the sequence limits (cf. the definition above) only depend on the behaviour of f above respectively below 1. Example: Consider $f(x)=\begin{cases}0 &\text{if }x\geq 1\\2&\text{if }x<1\end{cases}$. Then for all $n\in\mathbb{N}$ $f(u_n)=0$ and $f(v_n)=2$.

  2. I am not sure what you mean by 'define', but if you just want an example see my answer to the first question.

  3. No - No assumptions are necessary. Looking at the definitions for the left- and the right-hand limits of $f$ at $1$ it is (hopefully) clear that (if these limits exist) $\lim_{x \to 1^+} f(x)=\lim_{n\rightarrow\infty} f(u_n)$ and $\lim_{x \to 1^-} f(x)=\lim_{n\rightarrow\infty} f(v_n)$. So they cannot be the same since $2\not=0$. If either $\lim_{x \to 1^+} f(x)$ or $\lim_{x \to 1^-} f(x)$ does not exist neither can the limit $\lim_{x \to 1} f(x)$ since both $0 < 1-x <\delta$ and $0 < x-1 <\delta$ imply that $0 < |1-x| <\delta$.

  4. Taking into account the example I suspect that the left- and right-hand limits are what is meant by 'lateral limits'.

  5. Yes, b) can be justified (taking my answer to 4. to be correct) as follows: (I'll only do the case of the left-hand limit, but the right-hand limit is very similar.)

    Claim: If the limit $\lim_{x \to 1^-} f(x)$ exists, it holds that $\lim_{x \to 1^-} f(x)=2$

    Proof (by contradiction) Assume that $\lim_{x \to 1^-} f(x)=L$ with $L\not=2$. Then for every $\epsilon>0$ there exists $\delta>0$ such that for all x with $0<1-x<\delta$: $|f(x)-L|<\epsilon$. Since $\forall n\in\mathbb{N}:v_n<1$ and $v_n\rightarrow 1$ (as $n\rightarrow\infty$) there exists $N_1\in\mathbb{N}$ such that for all $n>N_1$: $0<1-v_n=|v_n-1|<\epsilon$. Furthermore, since $\lim_{n \to \infty} f(v_n)=2$, there exists $N_2\in\mathbb{N}$ such that for all $n>N_2$: $|f(v_n)-2|<\epsilon$. Thus, letting $N:=\max\{N_1,N_2\}$, it holds that for all $n>N$: $|f(v_n)-2|<\epsilon$ and $|f(v_n)-L|<\epsilon$. But this implies (for any $n>N$) $|L-2|=|f(v_n)-2-(f(v_n)-L)|\leq|f(v_n)-2|+|-(f(v_n)-L)|<\epsilon+\epsilon=2\epsilon$. Since this holds for every $\epsilon>0$ it holds for $\epsilon=\frac{|L-2|}{4}>0$ (since by assumption $|L-2|>0$). But this implies $|L-2|<2\frac{|L-2|}{4}$ which is equivalent (since by assumption $|L-2|>0$) to $2<1$ which is a contradiction. So the assumption must have been false. qed

    It is however important to note that $f$ does not necessarily have lateral limits! A counterexample is given by $f(x)=\begin{cases}0&\text{if }x\in\{u_n|n\in\mathbb{N}\}\\ 2&\text{if }x\in\{v_n|n\in\mathbb{N}\}\\ -1&\text{otherwise}\end{cases}$

Edit As requested a (hopefully) simplified explanation of the definitions and the idea behind the proof in the answer to question 5). Note: While I'll try to stay formally accurate the following is merely an explanation and should no be used as a substitute for the actual definitions / proof and rather as helper to understand them better.

On metric spaces instead of the limit $\lim_{x\rightarrow a} f(x)$ one may look at any sequence $x_n\in\mathbb{R}$ such that $\lim_{n\rightarrow\infty}x_n=a$ and consider $\lim_{n\rightarrow\infty} f(x_n)$. If the limit $\lim_{x\rightarrow a} f(x)$ exists then so does $\lim_{n\rightarrow\infty} f(x_n)$. To arrive the converse it is not enough to look at one such sequence $(x_n)_{n\in\mathbb{N}}$ but instead one needs to consider

(1) $\lim_{n\rightarrow\infty}f(x_n)$

for all sequences $(x_n)_{n\in\mathbb{N}}$ with $\lim_{n\rightarrow\infty} x_n=a$. If (1) exists for all such sequences $(x_n)_{n\in\mathbb{N}}$ and has the same value, then $\lim_{x\rightarrow a} f(x)$ exists and has this value as well.

It therefore helpful to look at the Limit of a sequence first: This definition simply means that, ignoring a finite number of elements of the sequence, all other elements are arbitrarily close to the limit.

Left- and right-hand limit: The idea behind these limits is to try and define limits in points where the function is discontinuous (i.e. the 'normal' limit does not exist). As discussed above one may (when taking the limit of $f$ in $a$) look at the behaviour of $f(x_n)$ for sequences $x_n\rightarrow a$ (for $n\rightarrow\infty$). The left- and right-hand limits consider only sequences that satisfy the additional condition that they approach their limit either from below (in case of the left-hand limit) or above (in case of the right-hand limit) and are never above respectively below their limit. Thus only the behaviour of $f$ on one side (as opposed to both sided for the 'normal' limit) of the point in which the limit is taken matters. That is to calculate for example $\lim_{x\rightarrow a^{-}} f(x)$ the definition of f on $(-\infty,a]$ matters, but $f$ can be arbitrary on $(a,\infty)$ and no knowledge of the behaviour of $f$ on this interval is needed.

Limit: The condition for the limit simply says that if $x$ is very close to $a$ then and $L=\lim_{x\rightarrow a} f(x)$ then $f(x)$ is very close to $L$.´

Proof in the answer to Q5: The basic idea is to realise that $v_n$ is one of the sequences considered for the left-hand limit. The sequence $v_n$ gets arbitrarily close to $1$. Thus for every $\delta$, after ignoring finitely many (i.e. $N_1$ many) elements of the sequence $v_n$, $x=v_n$ will be close enough to $1$ such that (by definition of the left-hand limit) $f(x)$ is very close ($\epsilon$) to the left-hand limit. But this means that $f(v_n)$ gets arbitrarily close to the left-hand limit $L$ and thus (since we assumed this limit exists) $\lim_{n\rightarrow\infty} f(v_n)=L$.

  • $\begingroup$ If it's not asking too much, could you explain 5. and the text in the beginning (before question 1.) in a non-mathematical way? In other words, could you write it in plain english, please? I can understand the rest though. $\endgroup$ – Mark Read Jan 21 '17 at 18:50
  • $\begingroup$ I can certainly try even though the $\epsilon$-$\delta$ stuff can often seem very technical. I will likely update my answer soon. $\endgroup$ – Jonathan von Schroeder Jan 21 '17 at 19:05
  • $\begingroup$ I hope the above helps to clarify things (I know it's not fully plain english, but hopefully it makes things more clear nonetheless). If not feel free to ask again. $\endgroup$ – Jonathan von Schroeder Jan 21 '17 at 19:57
  • $\begingroup$ I also realised that my definition for the limit was incomplete (the assumption $x\not=a$ was missing). Sorry about any confusion this may have caused. $\endgroup$ – Jonathan von Schroeder Jan 21 '17 at 20:02

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