Question

The value of $\mathop {\lim }\limits_{x \to 1} \dfrac{{x + {x^2} + ...... + {x^n} - n}}{{x - 1}}$ is
$

  {\text{a}}{\text{. n}} \\
  {\text{b}}{\text{. }}\dfrac{{n + 1}}{2} \\
  {\text{c}}{\text{. }}\dfrac{{n\left( {n + 1} \right)}}{2} \\
  {\text{d}}{\text{. }}\dfrac{{n\left( {n - 1} \right)}}{2} \\
$

Answer 1

Hint: - Apply L’ Hospital’s Rule.

Given limit is
$\mathop {\lim }\limits_{x \to 1} \dfrac{{x + {x^2} + ...... + {x^n} - n}}{{x - 1}}$
Put$\left( {x = 1} \right)$, in this limit
$ \Rightarrow \dfrac{{1 + 1 + 1 + .................. + 1 - n}}{{1 - 1}}$
As we know sum of 1 up to n terms is equal to n.
$ \Rightarrow \dfrac{{n - n}}{{1 - 1}} = \dfrac{0}{0}$
So, at\[x = 1\], the limit is in form of \[\dfrac{0}{0}\]
So, apply L’ Hospital’s rule
So, differentiate numerator and denominator separately w.r.t.$x$
As we know differentiation of\[{{\text{x}}^n} = n{x^{n - 1}}\], and differentiation of constant term is zero.
\[ \Rightarrow \mathop {\lim }\limits_{x \to 1} \dfrac{{\dfrac{d}{{dx}}\left( {x + {x^2} + ...... + {x^n} - n} \right)}}{{\dfrac{d}{{dx}}\left( {x - 1} \right)}} \Rightarrow \mathop {\lim }\limits_{x \to 1} \dfrac{{1 + 2x + 3{x^2} + ........ + n{x^{n - 1}} - 0}}{{1 - 0}}\]
Now, put\[x = 1\],\[ \Rightarrow \dfrac{{1 + 2 + 3 + ...................... + n}}{1}\]
\[ \Rightarrow 1 + 2 + 3 + ...................... + n = \sum\limits_{r = 1}^n r \]
Now as we know sum of first natural numbers is \[\left( {{\text{i}}{\text{.e}}{\text{.}}\sum\limits_{r = 1}^n r = \dfrac{{n\left( {n + 1} \right)}}{2}} \right)\]
\[\mathop {\lim }\limits_{x \to 1} \dfrac{{x + {x^2} + ...... + {x^n} - n}}{{x - 1}} = \sum\limits_{r = 1}^n r = \dfrac{{n\left( {n + 1} \right)}}{2}\]
Hence option (c) is the correct answer .

Note: - In such types of questions the key concept we have to remember is that, whenever the limit comes in the form of \[\dfrac{0}{0}\] always apply L’ hospital’s rule, (i.e. differentiate numerator and denominator separately), and always remember the sum of first natural numbers then we will get the required answer.