Where does it approach?

Calculus Level 2

Evaluate the limit: lim ⁡ x → 1 x + x 2 + x 3 + ⋯ + x n − n x − 1 \large \lim_{x\rightarrow 1}\frac{x+x^2+x^3+\cdots+x^n-n}{x-1}

n 2 n^2 n n n 2 + n n^2+n n ( n + 1 ) 2 \frac{n(n+1)}{2} n 2 + 1 n \frac{n^2+1}{n}

Vilakshan Gupta by Vilakshan Gupta , 19, India

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2 solutions

Relevant wiki: L'Hopital's Rule - Basic

L = lim ⁡ x → 1 x + x 2 + x 3 + ⋯ + x n − n x − 1 A 0/0 case, L’H o ˆ pital’s rule applies. = lim ⁡ x → 1 1 + 2 x + 3 x 2 + ⋯ + n x n − 1 1 Differentiate up and down w.r.t. x = 1 + 2 + 3 + ⋯ + n = n ( n + 1 ) 2 \begin{aligned} L & = \lim_{x \to 1} \frac {x+x^2+x^3+\cdots + x^n - n}{x-1} & \small \color{#3D99F6} \text{A 0/0 case, L'Hôpital's rule applies.} \\ & = \lim_{x \to 1} \frac {1+2x+3x^2+\cdots + nx^{n-1}}1 & \small \color{#3D99F6} \text{Differentiate up and down w.r.t. }x \\ & = 1+2+3+\cdots +n \\ & = \boxed{\dfrac {n(n+1)}2} \end{aligned}

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Jason Martin
Nov 8, 2017

Use L'Hopital's Rule and you get the sum 1 + 2 + 3 + … + n 1+2+3+\ldots +n , which simplifies to n ( n + 1 ) 2 \frac{n(n+1)}{2}

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