Lim, sup, and inf

Calculus Level 2

Let $x_n$ be the sequence

$2,\ -2^{1/2},\ 2^{1/3},\ -2^{1/4},\ 2^{1/5},\ -2^{1/6},\ \ldots.$

What are the values of

$\inf x_n, \ \ \liminf_{n\to\infty} x_n, \ \ \lim_{n\to\infty} x_n, \ \ \limsup_{n\to\infty} x_n, \ \ \sup x_n \, ?$

Notation: In the choices, $\text{DNE}$ means "does not exist."

-1,0,\text{DNE},0,1

-1,-1,0,1,1

-2^{1/2}, 0, 0, 0, 2

-2^{1/2}, \text{DNE}, \text{DNE}, \text{DNE}, 2

-1,-1, \text{DNE}, 1, 1

-2^{1/2},-2^{1/2},\text{DNE},2,2

-2^{1/2}, -1, \text{DNE}, 1, 2

1 solution

Ferenets Roman
May 4, 2019

Inf is $-2^{1/2}$ cause sequence with even indexes is increasing, so first element is the least.Same approach for sup $x_n$ ,but for elements with odd indexes.Lim of $x_n$ doesn't exist cause sequence will bounce between -1 and 1 as $n\rightarrow\infty$ .And lim of subsequences is clearly -1 and 1 respectively, as lim of $a^{1/n}$ for a $>=1$ is 1.

Lim, sup, and inf

1 solution

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