I wish to find $\displaystyle \lim_{n \rightarrow \infty}\frac{n+1}{\sqrt{n}}$.

Here is what I did:

$1.$ Rewrite $\frac{n+1}{\sqrt{n}}$ to $(n+1) \cdot \frac{1}{\sqrt{n}}$

$2.$ Then I can apply limit laws to do the following $$ \lim_{n \rightarrow \infty}\left((n+1) \cdot \frac{1}{\sqrt{n}}\right) $$ $$ \lim_{n \rightarrow \infty}(n+1) \cdot \lim_{n \rightarrow \infty}\frac{1}{\sqrt{n}} = \infty \cdot 0 = 0 $$

However, if you go to sites like Symbolab or Wolfram, they do different steps and get a result that is different from mine. What am I doing wrong? And why is it wrong?

  • 13
    $\begingroup$ What makes you think that $\infty\times 0=0$? This is not true at all. I hope you agree that $n=n^2\times\frac{1}{n}$ does not tend to $0$. $\endgroup$
    – Mark
    Commented Sep 1, 2020 at 19:52
  • 3
    $\begingroup$ See indeterminate forms. $\endgroup$
    – V.G
    Commented Sep 1, 2020 at 19:53
  • 9
    $\begingroup$ You are not using the product rule for limits correctly as well. Limit of the product equals the product of the limits when all the limits involved "exist". $\endgroup$
    – Anurag A
    Commented Sep 1, 2020 at 19:53
  • $\begingroup$ @Mark I agree but I am trying to take a step by step approach to learn things better(it works better for me). Then I apply the product rule whereafter one part happens to go to 0 $\endgroup$
    – That Guy
    Commented Sep 1, 2020 at 19:55
  • 3
    $\begingroup$ @nichl nss The product rule works when both limits are finite, or in some very specific cases that involve infinite limits. (for example $\infty\times\infty=\infty$ is indeed true). But $\infty\times 0$ is a problematic limit. $\endgroup$
    – Mark
    Commented Sep 1, 2020 at 19:56

4 Answers 4


There are two problems in your attempt:

  1. $\lim a\cdot b=\lim a\cdot\lim b$ can only be used if the limits exist, and this is not the case;

  2. the expression $\infty\cdot0$ has no meaning as a function from $\Bbb R\times\Bbb R\to \Bbb R$ because $\infty$ is not a real number, and its product with a real cannot equal anything unless you first define what it is and how it behaves.

A correct solution is

$$\frac{n+1}{\sqrt n}=\sqrt n+\frac 1{\sqrt n}\ge\sqrt n$$ and the limit does not exist.

  • 1
    $\begingroup$ Actually the last limit exists and is $+\infty$. $\endgroup$
    – Angelo
    Commented Sep 1, 2020 at 20:21
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    $\begingroup$ @Angelo: it is a matter of convention. You may also say that the limit diverges. In any case, $\lim\cdots=\infty$ is a conventional notation, not a true equality. $\endgroup$
    – user65203
    Commented Sep 1, 2020 at 20:22
  • 4
    $\begingroup$ There are two problems in your answer. (1) You are missing brackets in "$\lim a·b$". This is not a matter of convention; if you claim that you do not need the brackets then the stuff on the other side of the equation is bogus. (2) What you claim is a correct solution is wrong, because you cannot claim an equality or the inequality between things that don't exist! $\endgroup$
    – user21820
    Commented Sep 2, 2020 at 6:34
  • 1
    $\begingroup$ I proposed a fix to the issue of the claim that $\infty$ is a number $\endgroup$ Commented Sep 3, 2020 at 16:32
  • 1
    $\begingroup$ @user21820 You are spreading misinformation. The fact that his notational convention does not require grouping symbols on $\lim{a\cdot{b}}$ does not imply his equation is wrong, it just means his convention is nonstandard. Also, I have no idea about what you are talking about in your second point, since $\sqrt{n}$ exists as a real number. The inequality is not between symbols that do not exist. $\endgroup$
    – Angel
    Commented Nov 18, 2021 at 14:56

You've used the product rule for limits incorrectly as mentioned in the comments.

Moreover we have $\lim_{x\rightarrow\ 0}\sin(x)=0$, but $\lim_{x\rightarrow\ 0}\frac{\sin(x)}{x}=1.$

Since $\frac{n+1}{\sqrt{n}}=\sqrt{n}+\frac{1}{\sqrt{n}}$, $\lim_{n\rightarrow \infty}\frac{1}{\sqrt{n}}=0$ and $\lim_{n\rightarrow\infty}\sqrt{n}=\infty$, it follows that $\lim_{n\rightarrow\infty}\frac{n+1}{\sqrt{n}}=\infty$.

Other methods include L'Hôpital's rule for example.


Short Version

The product law for limits has hypotheses which the asker has neglected to verify. A better approach is to bound the sequence from below, which gives $$ \frac{n+1}{\sqrt{n}} > \sqrt{n} \implies \lim_{n\to\infty} \frac{n+1}{\sqrt{n}} > \lim_{n\to\infty} \sqrt{n} = +\infty. $$

In More Detail

The "Limit Laws" are theorems, which means that they have hypotheses which must first be satisfied before they can be applied. For example, you have applied a product law, which states (in this context)

Theorem 1: If $\{a_n\}$ and $\{b_n\}$ are two sequences of real numbers and there are real numbers $L$ and $M$ such that $$ \lim_{n\to\infty} a_n = L \qquad\text{and}\qquad \lim_{n\to\infty} b_n = M, $$ then $$ \lim_{n\to\infty} (a_n\cdot b_n) = L\cdot M. $$

Note the hypotheses: specifically, both limits must exist; each sequence involved must converge to some real number. If you attempt to apply this theorem as you have, it fails because your sequences do not satisfy the hypotheses. In particular, there is no real number $L$ such that

$$ \lim_{n\to\infty} (n+1) = L; $$

the sequence $\{ n+1 \}$ diverges. You cannot apply the product law for limits here, since the hypotheses of that theorem are not satisfied.

Instead, you are better off simplifying your original sequence a little differently, then applying the following theorem:

Theorem 2: If $\{a_n\}$ and $\{b_n\}$ are two sequences and $a_n \ge b_n$ for all $n$, then

  1. if there is some real $M$ such that $\lim_{n\to\infty} b_n = M$, then either $$ \lim_{n\to\infty} a_n \ge L $$ or the sequence $\{a_n\}$ is unbounded above, i.e. it diverges to positive infinity.

  2. if $\{b_n\}$ is unbounded above, then $\{a_n\}$ is also unbounded above.

  3. if $\{b_n\}$ is unbounded below, then nothing can be said about $\{a_n\}$.

This could be stated a little less precisely as "if $a_n \ge b_n$ for all $n$, then $\lim_{n\to\infty} a_n \ge \lim_{n\to\infty} b_n$." Note that this statement is quite imprecise, because we have no guarantee that either of the limits involved even exists (as real numbers) and it doesn't really make sense to compare things which don't exist. By working in a slightly larger number system (the "extended real numbers", which include two extra elements: $\pm\infty$), the less precise statement can be made somewhat more precise—this is left as an exercise to the reader.

In any event, Theorem 2 may be applied as follows: if $n$ is positive, then $$ \frac{n+1}{\sqrt{n}} = \frac{n}{\sqrt{n}} + \frac{1}{\sqrt{n}} = \sqrt{n} + \frac{1}{\sqrt{n}}. $$ As $\sqrt{n}$ is positive for all positive $n$, it follows that $\frac{1}{\sqrt{n}} > 0$ for all $n$. Therefore, for all $n$, $$ \frac{n+1}{\sqrt{n}} = \sqrt{n} + \frac{1}{\sqrt{n}} > \sqrt{n}. $$ Note that $\sqrt{n}$ is unbounded above so, applying Theorem 2, conclude that the original sequence is unbounded above as well. In slightly less precise language $$ \frac{n+1}{\sqrt{n}} > \sqrt{n} \implies \lim_{n\to\infty} \frac{n+1}{\sqrt{n}} \ge \lim_{n\to\infty} \sqrt{n} = +\infty. $$


Answer :

$\lim_{n \to +\infty } \frac{n+1}{\sqrt{n}}$= $\frac{\sqrt{n}(\sqrt{n} +\frac{1}{\sqrt{n}})}{\sqrt{n}} $= $\lim_{n \to +\infty } \sqrt{n} +\frac{1}{\sqrt{n}}$ = $+\infty$


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