Which of the following limits is equal to $\int _ { 3 } ^ { 5 } x ^ { 4 } d x$ ?
(A) $\lim _ { n \rightarrow \infty } \sum _ { k = 1 } ^ { n } \left( 3 + \frac { k } { n } \right) ^ { 4 } \frac { 1 } { n }$
(B) $\lim _ { n \rightarrow \infty } \sum _ { k = 1 } ^ { n } \left( 3 + \frac { k } { n } \right) ^ { 4 } \frac { 2 } { n }$
(C) $\lim _ { n \rightarrow \infty } \sum _ { k = 1 } ^ { n } \left( 3 + \frac { 2 k } { n } \right) ^ { 4 } \frac { 1 } { n }$
(D) $\lim _ { n \rightarrow \infty } \sum _ { k = 1 } ^ { n } \left( 3 + \frac { 2 k } { n } \right) ^ { 4 } \frac { 2 } { n }$

Consider $f : \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$ defined as follows: $$f(a,b) := \lim_{n \rightarrow \infty} \frac{1}{n} \log_{e}\left[e^{na} + e^{nb}\right]$$
For each statement, state if it is true or false.
(a) $f$ is not onto i.e. the range of $f$ is not all of $\mathbb{R}$.
(b) For every $a$ the function $x \mapsto f(a,x)$ is continuous everywhere.
(c) For every $b$ the function $x \mapsto f(x,b)$ is differentiable everywhere.
(d) We have $f(0,x) = x$ for all $x \geq 0$.

(Calculus) A continuous function $f(x)$ defined on the closed interval $[0, 1]$ satisfies $f(0) = 0$, $f(1) = 1$, has a second derivative on the open interval $(0, 1)$, and $f'(x) > 0$, $f''(x) > 0$. Which of the following is equal to $\int_0^1 \{f^{-1}(x) - f(x)\} dx$? [3 points]
(1) $\lim_{n \rightarrow \infty} \sum_{k=1}^{n} \left\{\frac{k}{n} - f\left(\frac{k}{n}\right)\right\} \frac{1}{2n}$
(2) $\lim_{n \rightarrow \infty} \sum_{k=1}^{n} \left\{\frac{k}{n} - f\left(\frac{k}{n}\right)\right\} \frac{2}{n}$
(3) $\lim_{n \rightarrow \infty} \sum_{k=1}^{n} \left\{\frac{k}{n} - f\left(\frac{k}{n}\right)\right\} \frac{1}{n}$
(4) $\lim_{n \rightarrow \infty} \sum_{k=1}^{n} \left\{\frac{k}{2n} - f\left(\frac{k}{n}\right)\right\} \frac{1}{n}$
(5) $\lim_{n \rightarrow \infty} \sum_{k=1}^{n} \left\{\frac{2k}{n} - f\left(\frac{k}{n}\right)\right\} \frac{1}{n}$

There is a function $f ( x ) = x ^ { 2 } + a x + b \quad ( a \geqq 0 , b > 0 )$. For a natural number $n \geq 2$, divide the closed interval $[ 0,1 ]$ into $n$ equal parts, and let the division points (including both endpoints) be $$0 = x _ { 0 } , x _ { 1 } , x _ { 2 } , \cdots , x _ { n - 1 } , x _ { n } = 1$$ respectively. Let $A _ { k }$ be the area of the rectangle with base $\left[ x _ { k - 1 } , x _ { k } \right]$ and height $f \left( x _ { k } \right)$. $( k = 1,2 , \cdots , n )$
Given that the sum of the areas of the two rectangles at the ends is $$A _ { 1 } + A _ { n } = \frac { 7 n ^ { 2 } + 1 } { n ^ { 3 } }$$ find the value of $\lim _ { n \rightarrow \infty } \sum _ { k = 1 } ^ { n } \frac { 8 k } { n } A _ { k }$. [4 points]

For the function $f ( x ) = \frac { 1 } { x }$, what is the value of $\lim _ { n \rightarrow \infty } \sum _ { k = 1 } ^ { n } f \left( 1 + \frac { 2 k } { n } \right) \frac { 2 } { n }$? [3 points]
(1) $\ln 6$
(2) $\ln 5$
(3) $2 \ln 2$
(4) $\ln 3$
(5) $\ln 2$

For the function $f ( x ) = 4 x ^ { 3 } + x$, what is the value of $\lim _ { n \rightarrow \infty } \sum _ { k = 1 } ^ { n } \frac { 1 } { n } f \left( \frac { 2 k } { n } \right)$? [3 points]
(1) 6
(2) 7
(3) 8
(4) 9
(5) 10

What is the value of $\lim _ { n \rightarrow \infty } \frac { 1 } { n } \sum _ { k = 1 } ^ { n } \sqrt { \frac { 3 n } { 3 n + k } }$? [3 points]
(1) $4 \sqrt { 3 } - 6$
(2) $\sqrt { 3 } - 1$
(3) $5 \sqrt { 3 } - 8$
(4) $2 \sqrt { 3 } - 3$
(5) $3 \sqrt { 3 } - 5$

Show that $\forall x \in \mathbb { R } \backslash \{ - 1,1 \}$ $$\int _ { 0 } ^ { 2 \pi } \ln \left( x ^ { 2 } - 2 x \cos \theta + 1 \right) \mathrm { d } \theta = \lim _ { n \rightarrow + \infty } \left( \frac { 2 \pi } { n } \sum _ { k = 1 } ^ { n } \ln \left( x ^ { 2 } - 2 x \cos \frac { 2 k \pi } { n } + 1 \right) \right)$$

Consider the function $h : ]0,1[ \longrightarrow \mathbf{R},\; t \longmapsto \frac{1}{\sqrt{t(1-t)}}$. Prove that: $$\lim_{n \longrightarrow +\infty} \sum_{k=1}^{2n-1} \frac{1}{2n} h\!\left(\frac{k}{2n}\right) = \int_0^1 h(t)\, dt.$$

Consider the function $h : ]0,1[ \longrightarrow \mathbf{R},\; t \longmapsto \frac{1}{\sqrt{t(1-t)}}$, and let $\tilde{h}$ denote its restriction to $\left]0, \frac{1}{2}\right]$. Show that: $$\lim_{n \rightarrow +\infty} \sum_{k=1}^{n} \frac{1}{2n+1} h\!\left(\frac{k}{2n+1}\right) = \int_0^{\frac{1}{2}} h(t)\, dt.$$ Deduce that: $$\lim_{n \rightarrow +\infty} \sum_{k=1}^{2n} \frac{1}{2n+1} h\!\left(\frac{k}{2n+1}\right) = \int_0^1 h(t)\, dt.$$

Deduce the limit: $$\lim_{n \rightarrow +\infty} \sum_{i=1}^{n-1} \frac{1}{\sqrt{i(n-i)}}.$$

Evaluate $\lim_{n \rightarrow \infty} \left\{\frac{1}{n+1} + \frac{1}{n+2} + \ldots + \frac{1}{n+n}\right\}$.

The value of $\lim_{n \to \infty} \sum_{r} \frac{6n}{9n^{2} - r^{2}}$ is
(a) 0
(b) $\log(3/2)$
(c) $\log(2/3)$
(d) $\log(2)$

If $a _ { n } = \left( 1 + \frac { 1 } { n ^ { 2 } } \right) \left( 1 + \frac { 2 ^ { 2 } } { n ^ { 2 } } \right) ^ { 2 } \left( 1 + \frac { 3 ^ { 2 } } { n ^ { 2 } } \right) ^ { 3 } \cdots \left( 1 + \frac { n ^ { 2 } } { n ^ { 2 } } \right) ^ { n }$, then $$\lim _ { n \rightarrow \infty } a _ { n } ^ { - 1 / n ^ { 2 } }$$ is
(A) 0
(B) 1
(C) $e$
(D) $\sqrt { e } / 2$

For each positive integer $n$, define a function $f_n$ on $[0,1]$ as follows: $$f_n(x) = \left\{ \begin{array}{ccc} 0 & \text{if} & x = 0 \\ \sin\frac{\pi}{2n} & \text{if} & 0 < x \leq \frac{1}{n} \\ \sin\frac{2\pi}{2n} & \text{if} & \frac{1}{n} < x \leq \frac{2}{n} \\ \sin\frac{3\pi}{2n} & \text{if} & \frac{2}{n} < x \leq \frac{3}{n} \\ \vdots & \vdots & \vdots \\ \sin\frac{n\pi}{2n} & \text{if} & \frac{n-1}{n} < x \leq 1 \end{array} \right.$$ Then, the value of $\lim_{n \rightarrow \infty} \int_0^1 f_n(x) dx$ is
(A) $\pi$
(B) 1
(C) $\frac{1}{\pi}$
(D) $\frac{2}{\pi}$

For each positive integer $n$, define a function $f _ { n }$ on $[ 0, 1 ]$ as follows: $$f _ { n } ( x ) = \left\{ \begin{array} { c c c } 0 & \text { if } & x = 0 \\ \sin \frac { \pi } { 2 n } & \text { if } & 0 < x \leq \frac { 1 } { n } \\ \sin \frac { 2 \pi } { 2 n } & \text { if } & \frac { 1 } { n } < x \leq \frac { 2 } { n } \\ \sin \frac { 3 \pi } { 2 n } & \text { if } & \frac { 2 } { n } < x \leq \frac { 3 } { n } \\ \vdots & \vdots & \vdots \\ \sin \frac { n \pi } { 2 n } & \text { if } & \frac { n - 1 } { n } < x \leq 1 \end{array} \right.$$ Then, the value of $\lim _ { n \rightarrow \infty } \int _ { 0 } ^ { 1 } f _ { n } ( x ) d x$ is
(A) $\pi$
(B) 1
(C) $\frac { 1 } { \pi }$
(D) $\frac { 2 } { \pi }$

What is the limit of $\sum _ { k = 1 } ^ { n } \frac { e ^ { - k / n } } { n }$ as $n$ tends to $\infty$ ?
(A) The limit does not exist.
(B) $\infty$
(C) $1 - e ^ { - 1 }$
(D) $e ^ { - 0.5 }$

If $a _ { n } = \left( 1 + \frac { 1 } { n ^ { 2 } } \right) \left( 1 + \frac { 2 ^ { 2 } } { n ^ { 2 } } \right) ^ { 2 } \left( 1 + \frac { 3 ^ { 2 } } { n ^ { 2 } } \right) ^ { 3 } \cdots \left( 1 + \frac { n ^ { 2 } } { n ^ { 2 } } \right) ^ { n }$, then $$\lim _ { n \rightarrow \infty } a _ { n } ^ { - 1 / n ^ { 2 } }$$ is
(a) 0 .
(B) 1 .
(C) $e$.
(D) $\sqrt { e } / 2$.

For $a \in \mathbb{R}$, $|a| > 1$, let $$\lim_{n\rightarrow\infty}\left(\frac{1 + \sqrt[3]{2} + \cdots + \sqrt[3]{n}}{n^{7/3}\left(\frac{1}{(an+1)^2} + \frac{1}{(an+2)^2} + \cdots + \frac{1}{(an+n)^2}\right)}\right) = 54$$
Then the possible value(s) of $a$ is/are
(A) $-9$
(B) $-6$
(C) $7$
(D) $8$

Let $f : ( 0,2 ) \rightarrow R$ be defined as $f ( x ) = \log _ { 2 } \left( 1 + \tan \left( \frac { \pi x } { 4 } \right) \right)$. Then, $\lim _ { n \rightarrow \infty } \frac { 2 } { n } \left( f \left( \frac { 1 } { n } \right) + f \left( \frac { 2 } { n } \right) + \ldots + f ( 1 ) \right)$ is equal to $\_\_\_\_$.

If $f : R \rightarrow R$ is given by $f ( x ) = x + 1$, then the value of $\lim _ { n \rightarrow \infty } \frac { 1 } { n } \left[ f ( 0 ) + f \left( \frac { 5 } { n } \right) + f \left( \frac { 10 } { n } \right) + \ldots + f \left( \frac { 5 ( n - 1 ) } { n } \right) \right]$ is:
(1) $\frac { 3 } { 2 }$
(2) $\frac { 5 } { 2 }$
(3) $\frac { 1 } { 2 }$
(4) $\frac { 7 } { 2 }$

$\lim _ { n \rightarrow \infty } \left( \frac { n ^ { 2 } } { \left( n ^ { 2 } + 1 \right) ( n + 1 ) } + \frac { n ^ { 2 } } { \left( n ^ { 2 } + 4 \right) ( n + 2 ) } + \frac { n ^ { 2 } } { \left( n ^ { 2 } + 9 \right) ( n + 3 ) } + \ldots + \frac { n ^ { 2 } } { \left( n ^ { 2 } + n ^ { 2 } \right) ( n + n ) } \right)$ is equal to
(1) $\frac { \pi } { 8 } + \frac { 1 } { 4 } \ln 2$
(2) $\frac { \pi } { 4 } + \frac { 1 } { 8 } \ln 2$
(3) $\frac { \pi } { 4 } - \frac { 1 } { 8 } \ln 2$
(4) $\frac { \pi } { 8 } + \ln \sqrt { 2 }$

$\lim_{n \rightarrow \infty} \frac{1}{2^n} \left( \frac{1}{\sqrt{1 - \frac{1}{2^n}}} + \frac{1}{\sqrt{1 - \frac{2}{2^n}}} + \frac{1}{\sqrt{1 - \frac{3}{2^n}}} + \ldots + \frac{1}{\sqrt{1 - \frac{2^n - 1}{2^n}}} \right)$ is equal to
(1) $\frac{1}{2}$
(2) 1
(3) 2
(4) $-2$

The value of $\lim _ { n \rightarrow \infty } \sum _ { k = 1 } ^ { n } \frac { n ^ { 3 } } { \left( n ^ { 2 } + k ^ { 2 } \right) \left( n ^ { 2 } + 3 k ^ { 2 } \right) }$ is :
(1) $\frac { ( 2 \sqrt { 3 } + 3 ) \pi } { 24 }$
(2) $\frac { 13 \pi } { 8 ( 4 \sqrt { 3 } + 3 ) }$
(3) $\frac { 13 ( 2 \sqrt { 3 } - 3 ) \pi } { 8 }$
(4) $\frac { ( 2 \sqrt { 3 } - 3 ) \pi } { 24 }$

Find the value of the limit $\lim _ { n \rightarrow \infty } \frac { 10 ^ { 10 } } { n ^ { 10 } } \left[ 1 ^ { 9 } + 2 ^ { 9 } + 3 ^ { 9 } + \cdots + ( 2 n ) ^ { 9 } \right]$ .
(1) $10 ^ { 9 }$
(2) $10 ^ { 9 } \times \left( 2 ^ { 10 } - 1 \right)$
(3) $2 ^ { 9 } \times \left( 10 ^ { 10 } - 1 \right)$
(4) $10 ^ { 9 } \times 2 ^ { 10 }$
(5) $2 ^ { 9 } \times 10 ^ { 10 }$