search
Log In
0 votes
85 views

The limit of the sequence $\sqrt{2}, \sqrt{2\sqrt{2}}, \sqrt{2\sqrt{2\sqrt{2}}}, \dots$ is

  1. $1$
  2. $2$
  3. $2\sqrt{2}$
  4. $\infty$
in Calculus
recategorized by
85 views
0
$2$ is the answer.

Please log in or register to answer this question.

Related questions

0 votes
1 answer
1
70 views
The value of $\underset{n \to \infty}{\lim} \bigg( \dfrac{1}{1-n^2} + \dfrac{2}{1-n^2} + \dots + \dfrac{n}{1-n^2} \bigg)$ is $0$ $ – \frac{1}{2}$ $\frac{1}{2}$ none of these
asked Sep 18, 2019 in Calculus gatecse 70 views
1 vote
1 answer
2
84 views
If $2f(x)-3f(\frac{1}{x})=x^2 \: (x \neq0)$, then $f(2)$ is $\frac{2}{3}$ $ – \frac{3}{2}$ $ – \frac{7}{4}$ $\frac{5}{4}$
asked Sep 18, 2019 in Calculus gatecse 84 views
0 votes
1 answer
3
126 views
Let $f(x) = \dfrac{x-1}{x+1}, \: f^{k+1}(x)=f\left(f^k(x)\right)$ for all $k=1, 2, 3, \dots , 99$. Then $f^{100}(10)$ is $1$ $10$ $100$ $101$
asked Sep 18, 2019 in Calculus gatecse 126 views
0 votes
2 answers
4
90 views
If $f(x)=e^{5x}$ and $h(x)=f’’(x)+2f’(x)+f(x)+2$ then $h(0)$ equals $38$ $8$ $4$ $0$
asked Sep 18, 2019 in Calculus gatecse 90 views
...