# 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} \\

$

Last updated date: 15th Mar 2023

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Answer

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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.

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.

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