4
$\begingroup$

How do I solve limits such as these? The $...$ always make it seem hard to me. From what I can understand from them, they both are $0/0$ limits, and I should be looking to write the numerator in such a way that the denominator should simplify it somehow. For the first I tried writing each element from the numerator as $(a-1)$ where $a = 1, x, x^2 ... x^n$ so that I can simplify the denominator eventually, but I didn't get much out of that. I'm guessing these two are solved in similar ways, hence why I posted them both. Any clues/hints I can get would be appreciated.

$$\lim \:_{x\to \:1}\frac{1+x+x^2+...\:+x^n-\left(n+1\right)}{x-1}$$ $$\lim _{x\to 1}\frac{x+x^2+...\:x^n-n}{x+x^2+...\:+x^m-m}$$

$n,\:m\:\in \mathbb{R}$

$\endgroup$
2
  • 1
    $\begingroup$ Hint: Write the numerator as $(1-1)+(x-1)+(x^2-1)+\dots+(x^n-1)$. Use the fact that $x^k-1 = (x-1)(x^{k-1}+x^{k-2}+\dots+x+1)$. $\endgroup$
    – Joey Zou
    Nov 25, 2015 at 16:58
  • 1
    $\begingroup$ You could just apply L'hospitals. $\endgroup$ Nov 25, 2015 at 16:59

6 Answers 6

3
$\begingroup$

Note that: $\frac{1+x+x^2+...\:+x^n-\left(n+1\right)}{x-1}= \frac{(x-1)+(x^2-1)+\ldots +(x^n-1)}{x-1}=\frac{(x-1)[1+(x+1)+(x^2+x+1)\ldots +(x^{n-1}+x^{n-2}+\ldots+x+1)]}{x-1}$,

after simplification we find:

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

So, $\lim \:_{x\to \:1}\frac{1+x+x^2+...\:+x^n-\left(n+1\right)}{x-1}=n+(n-1)+(n-2)+\ldots+2+1=\frac{n(n+1)}{2}$.

For the second one, we use the following method:

$\lim _{x\to 1}\frac{x+x^2+...\:x^n-n}{x+x^2+...\:+x^m-m}=\lim _{x\to 1}\frac{x+x^2+...\:x^n-n}{x-1}\times\frac{x-1}{x+x^2+...\:+x^m-m}=\frac{n(n+1)}{2}\times\frac{2}{m(m+1)}=\frac{n(n+1)}{m(m+1)}$

$\endgroup$
3
$\begingroup$

L'Hopital gives the answer instantly, but let's try to not use it.

The limit is just $$\sum_{i=1}^n \lim_{x \to 1} \frac{x^i-1}{x-1}$$ So it is $$\sum_{i=1}^n \lim_{x \to 1} (1+x+x^2+ \cdots x^{i-1}) = \sum_{i=1}^n i = \frac{n(n+1)}{2}$$

L'Hopital gives the answer immediatly since $$\lim_{x \to 1} \frac{1+x+ \cdots x^n-(n+1)}{x-1}=\lim_{x\to 1} \frac{1+2x+\cdots nx^{n-1}}{1} = 1+2+\cdots n = \frac{n(n+1)}{2}$$

$\endgroup$
3
$\begingroup$

For the first :

Let $f(x)=1+x+x^2+\cdots +x^n$. Then, we have $$\lim_{x\to 1}\frac{f(x)-f(1)}{x-1}=f'(1)$$

For the second :

Dividing the top and the bottom by $x-1$ gives $$\lim_{x\to 1}\dfrac{\dfrac{x+x^2+\cdots x^n-n}{x-1}}{\dfrac{x+x^2+\cdots +x^m-m}{x-1}}=\lim_{x\to 1}\frac{\frac{g(x)-g(1)}{x-1}}{\frac{h(x)-h(1)}{x-1}}=\frac{g'(1)}{h'(1)}$$ where $g(x)=x+x^2+\cdots x^n-n,h(x)=x+x^2+\cdots +x^m-m$.

$\endgroup$
0
$\begingroup$

Write this expression as $$\lim_{x \to 1}{x^n - 1\over {x - 1}}+{x^{n-1} - 1\over {x - 1}} + ... {x - 1\over {x-1}} + {1-1\over {x-1}}$$ You may then compute the limit using L'Hospital rule or using your basic knowledge of calculus.

$\endgroup$
0
$\begingroup$

HINT: $$\lim_{x\to \:1}\frac{1+x+x^2+...\:+x^n-\left(n+1\right)}{x-1} =^{L'H} \lim_{x\to1}\frac {1+2x+3x^2+...+nx^{n-1}}{1}$$

$\endgroup$
0
$\begingroup$

Use Hopital's rule, that leads to $ n(n+1)/2$.if you aren't familiar with it see https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule

$\endgroup$
2
  • $\begingroup$ I think your initial $1$ is wrong: $\frac{d}{dx}1 = 0$. $\endgroup$
    – TonyK
    Nov 25, 2015 at 17:25
  • $\begingroup$ Yes that's right.corrected! $\endgroup$
    – MAh2014
    Nov 25, 2015 at 17:29

Not the answer you're looking for? Browse other questions tagged .