Question

Exercise 2.3.1 Let x n

​
≥0 for all n∈N. (a) If (x n
​
)→0, show that ( x n
​
​
)→0. (b) If (x n
​
)→x, show that ( x n
​
​
)→ x
​
.

Answer 1

For a sequence xₙ ≥ 0 for all n ∈ N

a)
`If \( (x_n) \to 0 \)`, then
`\( (√(x_n)) \to 0 \)`.

(b) If
`\( ({x_n}) \to x \)` , then
`\( (x_n^2) \to x^2 \).`

Let's prove both parts:

`(a) If \( (x_n) \to 0 \),`

show that
`\( (√(x_n)) \to 0 \):`

Given
`\( (x_n) \to 0 \),` we want to show that
`\( (√(x_n)) \to 0 \).`

For any
`\( \epsilon > 0 \),` there exists N

such that for all
`( n > N \), \( |x_n - 0| < \epsilon^2 \) (since \( (x_n) \to 0 \)).`

Now, observe that

`\( |√(x_n) - 0| = √(x_n) < \epsilon \) for all \( n > N \)`

because
`\( \epsilon^2 \) is smaller than \( \epsilon \).`Therefore,
`\( (√(x_n)) \to 0 \).`

(b)
`If \( (x_n) \to x \),` show that
`\( (x_n^2) \to x^2 \):`

Given
`\( (x_n) \to x \),`

we want to show that
`\( (x_n^2) \to x^2 \).` For any
`\( \epsilon > 0 \),`

there exists
`\( N \)` such that for all
`\( n > N \), \( |x_n - x| < \epsilon \).`

Now, consider
`\( |x_n^2 - x^2| = |x_n - x| \cdot |x_n + x| \).` Since
`\( (x_n) \to x \), \( |x_n - x| \)`

can be made arbitrarily small for

`\( n > N \). Also, \( |x_n + x| \)` is bounded (it does not tend to infinity). Therefore,
`\( |x_n^2 - x^2|` trarily small for
`\( n > N \), and \( (x_n^2) \to x^2 \).`

These proofs use the properties of limits and the fact that the square root and square functions are continuous.