Induction Limit

Splitting n into 1 n times we have $\lim_{x \to 1} \frac {x^n - 1 + x^{n-1}- 1 + \dots + x- 1}{x-1}$ Which gives $\lim_{x \to 1} \frac {x^n - 1}{x-1}+\frac {x^{n-1}-1}{x-1} + \dots +\frac {x-1}{x-1}$ Then using the standard form $\lim_{x\to a} \frac {x^n-x^a}{x-a}=na^{n-1}$ we get the series of {n+(n-1)+(n-2)....1}; which is equal to $\frac {n(n+1)}{2}$

Both numerator and denominator were 0 so 0/0.

Applying L'Hospital Theorem and then replacing 1 will give

$\sum_{i=1}^3 i$

ans : n(n+1)/2

OR

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

dividing we get

$( x^{n-1} + x^{n-2} + ... + 1) + ( x^{n-2} + x^{n-3} + ... + 1) + ... + 1$

putting x = 1

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

ANS : ans : n(n+1)/2

Let use binomial expansion formulae

Let $\displaystyle x=1+\delta$

$\displaystyle \lim_{x \rightarrow 1} \dfrac{\Sigma_{i=1}^{n} x^{i}-n}{x-1}=\lim_{\delta \rightarrow 0}\dfrac{\Sigma_{i=1}^{n} (1+\delta)^{i}-n}{\delta}$

$\displaystyle (1+\delta)^{i}=1+i\delta$

$\displaystyle \lim_{\delta \rightarrow 0} \dfrac{n+\delta \dfrac{n(n+1)}{2}-n}{\delta}$

$\displaystyle \lim_{x \rightarrow 1} \dfrac{\Sigma_{i=1}^{n} x^{i}-n}{x-1}=\boxed{\dfrac{n(n+1)}{2}}$