_ n , ( n^2 - n + 1 n^2 - n - 1 )^ n(n - 1) = - Vidyadip

MCQ

$\mathop {\lim }\limits_{n \to \infty } \,{\left( {\frac{{{n^2} - n + 1}}{{{n^2} - n - 1}}} \right)^{n(n - 1)}} = $

A
$e$
✓
${e^2}$
C
${e^{ - 1}}$
D
$1$

✓

Answer

Correct option: B.

${e^2}$

b
(b) $\mathop {\lim }\limits_{n \to \infty } {\left( {\frac{{{n^2} - n + 1}}{{{n^2} - n - 1}}} \right)^{n(n - 1)}} = \mathop {\lim }\limits_{n \to \infty } {\left( {\frac{{n(n - 1) + 1}}{{n(n - 1) - 1}}} \right)^{n(n - 1)}}$

$ = \mathop {\lim }\limits_{n \to \infty } \frac{{{{\left( {1 + \frac{1}{{n(n - 1)}}} \right)}^{n(n - 1)}}}}{{{{\left( {1 - \frac{1}{{n(n - 1)}}} \right)}^{n(n - 1)}}}}$$ = \frac{e}{{{e^{ - 1}}}} = {e^2}$.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from STD 11 - 12. limits→STD 11 - 12. limits — all question types→Maths STD 11 Science question papers→CBSE Board STD 11 Science question papers→CBSE Board question paper generator→

Similar questions

Two dice are thrown simultaneously. Find the probability of getting an even number as the sum.

If the coefficients of $x^2$ and $x^3$ in the expansion of $(3+\text{ax})^{9}$ are the same, then the value of $a$ is:

The fourth term in the expansion of ${(1 - 2x)^{3/2}}$ will be

A point moves in the $x-y$ plane such that the sum of its distances from two mutually perpendicular lines is always equal to $3$. The area enclosed by the locus of the point is- ............... $\mathrm{unit}^{2}$

The variance of $20$ observation is $5$ . If each observation is multiplied by $2$ , then the new variance of the resulting observations, is

In the expansion of ${\left( {\frac{{3{x^2}}}{2} - \frac{1}{{3x}}} \right)^9}$,the term independent of $x$ is

By mathamatical induction $n(n^2− 1)$ is divisible by:

$\mathop {\lim }\limits_{x \to \pi /2} (\sec \theta - \tan \theta ) = $

Let $A = \{1, 2, 3, 4, 5\}; B = \{2, 3, 6, 7\}$. Then the number of elements in $(A × B) \cap (B × A)$ is

${{\left( \frac{1+\cos \varphi +i\sin \varphi }{1+\cos \varphi -i\sin \varphi } \right)}^{n}}=$