Question

Find the value of following limit:
 $\mathop {\lim }\limits_{x \to 1} \dfrac{{x + {x^2}... + {x^n} - n}}{{x - 1}}$
$
  {\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- The given expression is in the form $\dfrac{0}{0}$. So, we can use L’Hopital’s Rule here to solve the limit. This will help us to solve the question.

Complete step-by-step answer:
We have to find the limit of $\mathop {\lim }\limits_{x \to 1} \dfrac{{x + {x^2}... + {x^n} - n}}{{x - 1}}$.
When we put x = 1 in the above expression we get the expression to be of form $\dfrac{0}{0}$.
Now, as we know that if a limit is in the form $\dfrac{0}{0}$, then we can use L’Hopital’s Rule to solve the limit.
So, applying use L’Hopital’s Rule on the given Equation we get,
$\mathop {\lim }\limits_{x \to 1} \dfrac{{x + {x^2}... + {x^n} - n}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \dfrac{{1 + 2x... + n{x^{n - 1}}}}{1}$
$\mathop { \Rightarrow \lim }\limits_{x \to 1} \dfrac{{x + {x^2}... + {x^n} - n}}{{x - 1}} = \dfrac{{1 + 2... + n}}{1}$
$\mathop { \Rightarrow \lim }\limits_{x \to 1} \dfrac{{x + {x^2}... + {x^n} - n}}{{x - 1}} = 1 + 2 + ..... + n$
Now 1 + 2 + … + n is in A.P.
So, the Sum of A.P. is ${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$
Using this with a = 1, d = 1, we get,
${S_n} = \dfrac{n}{2}\left( {2 \cdot 1 + \left( {n - 1} \right)1} \right)$
$ \Rightarrow {S_n} = \dfrac{n}{2}\left( {2 + n - 1} \right)$
$ \Rightarrow {S_n} = \dfrac{{n\left( {n + 1} \right)}}{2}$
Therefore, $\mathop { \Rightarrow \lim }\limits_{x \to 1} \dfrac{{x + {x^2}... + {x^n} - n}}{{x - 1}} = \dfrac{{n\left( {n + 1} \right)}}{2}$
Hence, option C is correct.

Note- Whenever we face such types of problems the main point to remember is that we need to have a good grasp over limits and differentiation. In these types of questions, we should always use L’Hopital’s Rule. If it is applicable on the expression This helps in getting us the required condition and gets us on the right track to reach the answer.