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Recent questions tagged limits

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0 answers
1
The limit $\underset{n\rightarrow \infty }{\lim}\:n^{2}\int_{0}^{1}\:\frac{1}{\left ( 1+x^{2} \right )^{n}}\:dx$ is equal to $1$ $0$ $+\infty$ $1/2$
asked Aug 30, 2020 in Calculus soujanyareddy13 180 views
2 votes
1 answer
3
Let $a_n=\bigg( 1 – \frac{1}{\sqrt{2}} \bigg) \cdots \bigg( 1 – \frac{1}{\sqrt{n+1}} \bigg), \: n \geq 1$. Then $\underset{n \to \infty}{\lim} a_n$ equals $1$ does not exist equals $\frac{1}{\sqrt{\pi}}$ equals $0$
asked Sep 23, 2019 in Calculus Arjun 275 views
4 votes
4 answers
4
$\underset{x \to \infty}{\lim} \left( \frac{3x-1}{3x+1} \right) ^{4x}$ equals $1$ $0$ $e^{-8/3}$ $e^{4/9}$
asked Sep 23, 2019 in Calculus Arjun 420 views
3 votes
2 answers
5
$\underset{n \to \infty}{\lim} \dfrac{1}{n} \bigg( \dfrac{n}{n+1} + \dfrac{n}{n+2} + \cdots + \dfrac{n}{2n} \bigg)$ is equal to $\infty$ $0$ $\log_e 2$ $1$
asked Sep 23, 2019 in Calculus Arjun 301 views
2 votes
2 answers
6
$\underset{x \to 2}{\lim} \dfrac{1}{1+e^{\frac{1}{x-2}}}$ is $0$ $1/2$ $1$ non-existent
asked Sep 23, 2019 in Calculus Arjun 163 views
0 votes
0 answers
7
Let $f(x)$ be a continuous function from $[0,1]$ to $[0,1]$ satisfying the following properties. $f(0)=0$, $f(1)=1$, and $f(x_1)<f(x_2)$ for $x_1 < x_2$ with $0 < x_1, \: x_2<1$. Then the number of such functions is $0$ $1$ $2$ $\infty$
asked Sep 23, 2019 in Calculus Arjun 169 views
1 vote
2 answers
8
Let $f: \bigg( – \dfrac{\pi}{2}, \dfrac{\pi}{2} \bigg) \to \mathbb{R}$ be a continuous function, $f(x) \to +\infty$ as $x \to \dfrac{\pi^-}{2}$ and $f(x) \to – \infty$ as $x \to -\dfrac{\pi^+}{2}$. Which one of the following functions satisfies the above properties of $f(x)$? $\cos x$ $\tan x$ $\tan^{-1} x$ $\sin x$
asked Sep 23, 2019 in Calculus Arjun 137 views
0 votes
0 answers
9
Let $f(x) = \begin{cases}\mid \:x \mid +1, & \text{ if } x<0 \\ 0, & \text{ if } x=0 \\ \mid \:x \mid -1, & \text{ if } x>0. \end{cases}$ Then $\underset{x \to a}{\lim} f(x)$ exists if $a=0$ for all $a \in R$ for all $a \neq 0$ only if $a=1$
asked Sep 23, 2019 in Calculus Arjun 97 views
0 votes
0 answers
10
$\underset{x \to 0}{\lim} \dfrac{x \tan x}{1- \cos tx}$ is equal to $0$ $1$ $\infty$ $2$
asked Sep 23, 2019 in Calculus Arjun 93 views
2 votes
2 answers
11
The value of the infinite product $P=\frac{7}{9} \times \frac{26}{28} \times \frac{63}{65} \times \cdots \times \frac{n^3-1}{n^3+1} \times \cdots \text{ is }$ $1$ $2/3$ $7/3$ none of the above
asked Sep 23, 2019 in Calculus Arjun 216 views
0 votes
1 answer
12
The limit $\:\:\:\underset{n \to \infty}{\lim} \Sigma_{k=1}^n \begin{vmatrix} e^{\frac{2 \pi i k }{n}} – e^{\frac{2 \pi i (k-1) }{n}} \end{vmatrix}\:\:\:$ is $2$ $2e$ $2 \pi$ $2i$
asked Sep 23, 2019 in Calculus Arjun 215 views
1 vote
1 answer
13
The limit $\underset{n \to \infty}{\lim} \left( 1- \frac{1}{n^2} \right) ^n$ equals $e^{-1}$ $e^{-1/2}$ $e^{-2}$ $1$
asked Sep 23, 2019 in Calculus Arjun 147 views
0 votes
1 answer
14
Let $a_n= \bigg( 1 – \frac{1}{\sqrt{2}} \bigg) \cdots \bigg( 1- \frac{1}{\sqrt{n+1}} \bigg), \: \: n \geq1$. Then $\underset{n \to \infty}{\lim} a_n$ equals $1$ does not exist equals $\frac{1}{\sqrt{\pi}}$ equals $0$
asked Sep 23, 2019 in Calculus Arjun 167 views
1 vote
1 answer
15
The limit $\displaystyle{}\underset{x \to \infty}{\lim} \left( \frac{3x-1}{3x+1} \right) ^{4x}$ equals $1$ $0$ $e^{-8/3}$ $e^{4/9}$
asked Sep 23, 2019 in Calculus Arjun 197 views
0 votes
0 answers
16
$\displaystyle{}\underset{n \to \infty}{\lim} \frac{1}{n} \bigg( \frac{n}{n+1} + \frac{n}{n+2} + \cdots + \frac{n}{2n} \bigg)$ is equal to $\infty$ $0$ $\log_e 2$ $1$
asked Sep 23, 2019 in Calculus Arjun 141 views
0 votes
0 answers
17
Let $\{a_n\}$ be a sequence of real numbers. Then $\underset{n \to \infty}{\lim} a_n$ exists if and only if $\underset{n \to \infty}{\lim} a_{2n}$ and $\underset{n \to \infty}{\lim} a_{2n+2}$ exists $\underset{n \to \infty}{\lim} a_{2n}$ ... $\underset{n \to \infty}{\lim} a_{3n}$ exist none of the above
asked Sep 23, 2019 in Calculus Arjun 160 views
1 vote
1 answer
18
Suppose $a>0$. Consider the sequence $a_n = n \{ \sqrt[n]{ea} – \sqrt[n]{a}, \:\:\:\:\: n \geq 1$. Then $\underset{n \to \infty}{\lim} a_n$ does not exist $\underset{n \to \infty}{\lim} a_n=e$ $\underset{n \to \infty}{\lim} a_n=0$ none of the above
asked Sep 23, 2019 in Calculus Arjun 125 views
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