Answer
64.8k+ views
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.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What is the common property of the oxides CONO and class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The area of square inscribed in a circle of diameter class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Other Pages
In the ground state an element has 13 electrons in class 11 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
According to classical free electron theory A There class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Electric field due to uniformly charged sphere class 12 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
A boat takes 2 hours to go 8 km and come back to a class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)