Finding value of $\lim_{x \to 1} \dfrac{(1-x^n)}{(1-x)}$ without using L'Hospital

Question

How do I find the value of $$\lim_{x \to 1} \dfrac{(1-x^n)}{(1-x)}$$

I know it could be done through L'Hospital, but I am struggling to find another solution.

Answer 1

You can simplify the expression by writting $(1 - x^n)$ as $(1 - x) f(x)$, where $f(x)$is equal to $(1 + x + x^2 + ... +x^{n - 1})$. Now you get:

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

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

Answer 2

$$ \begin{aligned} \lim _{x \rightarrow 1} \frac{1-x^n}{1-x} & =\lim _{x \rightarrow 1} \frac{x^n-1^n}{x-1} \\ & =\left.\frac{d}{d x}\left(x^n\right)\right|_{x=1} \quad \textrm{ (By the definition of derivative )} \\ & =\left.n x^{n-1}\right|_{x=1} \\ & =n \end{aligned} $$

Answer 3

Substitution $$ \begin{aligned} \lim _{x \rightarrow 1} \frac{1-x^n}{1-x} = & \lim _{y \rightarrow 0} \frac{1-(1-y)^n}{y}, \text { where } y-1-x \\ = & \lim _{y \rightarrow 0} \frac{1- \displaystyle \sum_{k=0}^n\left(\begin{array}{l} n \\ k \end{array}\right)(-y)^k}{y} \\ = & \lim _{y \rightarrow 0}\left(\sum_{k=1}^n\left(\begin{array}{l} n \\ k \end{array}\right)(-1)^{k+1} y^{k-1}\right) \\ = & \left(\begin{array}{l} n \\ 1 \end{array}\right)(-1)^2 \\ = & n \end{aligned} $$