MCQ
$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {\frac{1}{n}{e^{\frac{r}{n}}}} $ is
  • A
    $e + 1$
  • $e - 1$
  • C
    $1 - e$
  • D
    $e$

Answer

Correct option: B.
$e - 1$
b
(b) $\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {\frac{1}{n}{e^{\frac{r}{n}}} = \int_0^1 {{e^x}dx = [{e^x}]_0^1 = e - 1} } $.

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

Similar questions

A box contains $b$ blue balls and $r$ red balls. A ball is drawn randomly from the box and is returned to the box with another ball of the same colour. The probability that the second ball drawn from the box is blue, is
$x * y=x^{2}+y^{3}$ and $(x * 1) * 1=x *(1 * 1)$. Then a value of $2 \sin ^{-1}\left(\frac{x^{4}+x^{2}-2}{x^{4}+x^{2}+2}\right)$ is
${d \over {dx}}\left[ {{{\tan }^{ - 1}}\sqrt {{{1 - \cos x} \over {1 + \cos x}}} } \right] = $
If ${a^2} + 4{b^2} = 12ab,$ then $\log (a + 2b)$ is
Consider circle $S$ : $x^2 + y^2 = 1$ and $P(0, -1)$ on it. $A$ ray of light gets reflected from tangent to $S$ at $P$ from the point with abscissa $-3$ and becomes tangent to the circle $S.$  Equation of reflected ray is
$\int_0^{\pi /2} {{{\left( {\frac{\theta }{{\sin \theta }}} \right)}^2}d\theta = } $
If $\omega $ is a complex cube root of unity, then $(x - y)(x\omega - y)$ $(x{\omega ^2} - y) = $
If the sum of $n$ terms of an $A.P.$ is $nA + {n^2}B$, where $A,B$ are constants, then its common difference will be
Let $X$ be a random variable such that the probability function of a distribution is given by $P(X=$ 0) $=\frac{1}{2}, \mathrm{P}(\mathrm{X}=\mathrm{j})=\frac{1}{3^{j}}(\mathrm{j}=1,2,3, \ldots, \infty)$. Then the mean of the distribution and $\mathrm{P}(\mathrm{X}$ is positive and even) respectively are:
A car completes the first half of its journey with a velocity ${v_1}$ and the rest half with a velocity ${v_2}$. Then the average velocity of the car for the whole journey is