Limit of $\lim\limits_{x \rightarrow 1} \frac{ (x +x^2+x^3+ \cdots +x^n)-n}{x-1}$

Question

What is the limit of $$\lim\limits_{x \rightarrow 1} \frac{ (x +x^2+x^3+ \cdots +x^n)-n}{x-1}$$

Answer 1

Hint: use l'Hospital's rule. You will also need to know how to sum up the first $n$ positive integers.

Edit: Here is some more detail if you need it. $$\lim_{x\to 1}(x+x^2+\cdots+x^n-n)=n-n=0$$ And $$\lim_{x\to 1} (x-1)=0.$$ So we may use l'Hospital's rule to show that your limit is the same as the limit $$\lim_{x\to 1}(1+2x+\cdots+nx^{n-1}).$$ Can you see how to finish from here?

Answer 2

Notice that $1$ is a root of the numerator of this fraction, meaning $(x-1)$ will cleanly divide it. Using polynomial long division, you'll find:

$$\frac{x^n + x^{n-1} + \cdots + x - n}{x-1} = \sum_{k=1}^{n} kx^{n-k}$$

And you can now find the limit by plugging in $x=1$: no more division-by-zero.

Answer 3

It's has two ways: first by taking Hopital : because by setting $x=1$ we have $\frac{0}{0}$ for this fraction: Now I take Hopital( taking derivative of numerator and denominator separately):

$$\lim_{x \to 1}\frac{x+x^2+...+x^n-n}{x-1}=\lim_{x \to 1}\frac{1+2x+3x^2...+nx^{n-1}}{1}$$ now by setting $x=1$ we have:

$$Limit=\frac{n(n+1)}{2}$$

In second method, I use geometric series for summing numerator.

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

Now replace this in main limitation:

$$=\lim_{x \to 1}\frac{\frac{x(x^n-1)}{x-1}-n}{x-1}=\lim_{x \to 1}\frac{x^{n+1}-(n+1)x+n}{(x-1)^2}$$

No by setting x=1: we have $\frac{0}{0}$ for this limit and I take Hopital double times for resolving this matter.

first Hopital: $$=\lim_{x \to 1}\frac{(n+1)x^{n}-(n+1)}{2(x-1)}$$

first Hopital: $$=\lim_{x \to 1}\frac{n(n+1)x^{n-1}}{2}=\frac{n(n+1)}{2}$$

Notice: because we have ${x \to 1} $ so I could not use $\frac{a}{1-q}$ for calculating sumation of geometric series.