A problem related with limit and continuity.

Question

What is the limit as $x$ tends to $1$ in the function $$\frac{x+x^2+x^3+\cdots+x^n-n}{x-1}\quad?$$

I tried factoring out the $x$ term then again rewriting the factored term. But nothing came out at all. Could anyone help me out?

Please show your attempts to avoid closure of your question. — ə̷̶̸͇̘̜́̍͗̂̄︣͟, Aug 07 '18 at 14:41
note that $1$ is root of nominator so....$x+x^2+...-n=(x-1)P$ — dmtri, Aug 07 '18 at 14:50
dmtri.Why not write an answer with your suggestion, polynomial long division .Was just about to write an answer when I saw your comment.Greetings. — Peter Szilas, Aug 07 '18 at 16:27
@ Peter Szilas, please feel free to write your answer even if we had the same thoughts. Most of the times I do not write answers because latex is really painful through my android phone. I make comments though just in case the OP needs quickly a short hint... — dmtri, Aug 08 '18 at 11:44
dmtri.Also write my answers with my mobile.Also not exceptionally good at latex, also have to be careful with the spell checker , does funny things at times.So I am quite slow , never mind. If I post and the post is just too messy, I delete, and then there is plenty of time to correct:)Greetings. — Peter Szilas, Aug 08 '18 at 15:33

ə̷̶̸͇̘̜́̍͗̂̄︣͟ · Answer 1 · 2018-08-07T16:55:15.313

5

HINT:

Notice that you can factor $(x-1)$ from the numerator. This means that you can divide by $x-1$ and you are left with a straightforward plugging in of numbers. $$\sum_{k=1}^nx^k-n=(x-1)[1+(x+1)+(x^2+x+1)+\cdots+(x^{n-1}+\cdots+x+1)]$$

edited Aug 07 '18 at 16:55

answered Aug 07 '18 at 14:44

ə̷̶̸͇̘̜́̍͗̂̄︣͟

27,005

Thanks man. Didn't realise that. – S.Nep Aug 07 '18 at 14:46
You're welcome. As a more general tip, try to include what you've tried when asking a question :) – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Aug 07 '18 at 14:46
First, your parenthetical term is not correct; the last term should be $x^{n-1}$, not $x^n$. Second, how does using $\sum_{k=1}^n x^k=\frac{x-x^{n+1}}{1-x}$ facilitate evaluation of the limit of interest? – Mark Viola Aug 07 '18 at 16:43
@MarkViola Ah you're correct. I somehow thought that the $x-1$ terms would cancel out. I've deleted that part of my answer now. – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Aug 07 '18 at 16:55

Rushabh Mehta · Answer 2 · 2018-08-07T16:51:15.950

4

Rewrite the limit as the following:

$$\lim_{x\to1}\frac{x-1}{x-1} + \lim_{x\to1}\frac{x^2-1}{x-1} + ... + \lim_{x\to1}\frac{x^n-1}{x-1}$$

Notice that each limit term has an $x-1$ term in the numerator and denominator. Since in our limit as $x\to1$, $x\neq1$, we can divide through by $x-1$ in every term.

$$\lim_{x\to1}1 + \lim_{x\to1} (x+1) + ... \lim_{x\to1} (x^{n-1} + ...+x+ 1)$$

Since each limit now is of a continuous function with a defined value at $x=1$, we can drop the limits, and substitute $x=1$ to get

$$=\sum\limits_{i=1}^n i = \frac{(n+1)\cdot n}2$$

edited Aug 07 '18 at 16:51

answered Aug 07 '18 at 14:47

Rushabh Mehta

13,663

Actually, each term has a removable singularity at $x=1$. – Mark Viola Aug 07 '18 at 16:44
I mean, that's exactly what I exploited here – Rushabh Mehta Aug 07 '18 at 16:45
That is not, however, what is written. Each term is undefined at $1$, but has a removable discontinuity (i.e., The limits as $x\to 1$ exist.). – Mark Viola Aug 07 '18 at 16:46
Fine, I'll edit to make it more clear – Rushabh Mehta Aug 07 '18 at 16:49
(+1) Well done! – Mark Viola Aug 07 '18 at 20:16

score 3 · Answer 3 · answered Aug 07 '18 at 15:37

Define

$$f(x)=x+x^2+x^3+\cdots+x^n.$$

Then $$f'(x)=1+2x+3x^2+\cdots+nx^{n-1}.$$

It follows that \begin{align*} \lim_{x \to 1}\frac{x+x^2+x^3+\cdots+x^n-n}{x-1}&=\lim_{x \to 1}\frac{f(x)-f(1)}{x-1}=f'(1)\\&=1+2+3+\cdots+n\\&=\frac{n(n+1)}{2}.\end{align*}

score 2 · Answer 4 · answered Aug 07 '18 at 14:46

2

Hint:

Write $n$ as $1+1+1\cdots$ with $1$occuring $n$ times and group it with each of the $x,x^2,x^3,\cdots x^n$ term and factor out the $x-1$ term to get the limit.

Method 2: Use L'Hospital rule

answered Aug 07 '18 at 14:46

Rohan Shinde

9,737

Peter Szilas · Answer 5 · 2018-08-08T15:50:07.357

2

As mentioned by user dimitry in his comment:

$P(x)= x^n+x^{n-1}......+x -n;$

$P(x=1)= 0$, hence $(x-1)$ is a factor of $P(x)$.

Polynomial long division yields:

$F(x):=$

$ (x^n +x^{n-1}...+x -n) ÷ (x-1)= $

$x^{n-1} +2x^{n-2} +......(n-1)x + n$;

$\lim_{x \rightarrow 1} F(x)=$

$1+2+...(n-1)+n =$

$ \dfrac{n(n+1)}{2}.$

Ref.:https://en.m.wikipedia.org/wiki/Polynomial_long_division

edited Aug 08 '18 at 15:50

answered Aug 08 '18 at 15:38

Peter Szilas

20,344
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A problem related with limit and continuity.

5 Answers5