# \lim_{n \to \infty} \sqrt(n+1)/n

## Homework Statement

To find the limit of the sequence (I suppose it is zero)
$$\lim \frac{\sqrt{n+1}}{n}$$

## Homework Equations

The limit is zero if for every $\epsilon > 0$ there exists $n_0$ such that for all $n \geq n_0$ one has
$$|a_n - 0| < \epsilon$$
or what is the same
$$-\epsilon < a_n < +\epsilon$$

## The Attempt at a Solution

Tried writing it as
$$(n+1)^{1/2} n^{-1}$$
but have no idea what to do next.

I know I can think the associated differentiable function $$f(x) = \frac{\sqrt{x+1}}{x}$$ and use L'Hopital, but that would be cheating, as this concept is previous to the limit of functions.

Last edited by a moderator:

Related Calculus and Beyond Homework Help News on Phys.org
Mark44
Mentor

## Homework Statement

To find the limit of the sequence (I suppose it is zero)
$$\lim \frac{\sqrt{n+1}}{n}$$

## Homework Equations

The limit is zero if for every $\epsilon > 0$ there exists $n_0$ such that for all $n \geq n_0$ one has
$$|a_n - 0| < \epsilon$$
or what is the same
$$-\epsilon < a_n < +\epsilon$$

## The Attempt at a Solution

Tried writing it as
$$(n+1)^{1/2} n^{-1}$$
but have no idea what to do next.

I know I can think the associated differentiable function $$f(x) = \frac{\sqrt{x+1}}{x}$$ and use L'Hopital, but that would be cheating, as this concept is previous to the limit of functions.
You have $\frac{\sqrt{n+1}}{n} < \epsilon$
Start by multiplying both sides by n. Since n > 0, this won't change the direction of the inequality. Next, square both sides, which gives you a quadratic inequality that you can solve for n.

dextercioby
Homework Helper
You can use the double inequality

$$0 < \frac{\sqrt{1+n}}{n} <\frac{2}{\sqrt{1+n}}$$, valid for n>1.

Hi Mark44,
Thanks, that gives me

$$\alpha_+ = \frac{1 + \sqrt{1+4\epsilon^2}}{2\epsilon^2}$$
$$\alpha_- = \frac{1 - \sqrt{1+4\epsilon^2}}{2\epsilon^2}$$

So it suffices to take $$n_0 = [\alpha_+ + 1]$$ right?
(the least integer greater than the greater of the real roots)

(It works, still, I think there should be a nicer way, without feching roots of polynomials)

I have a related question: my teacher used in part of an excercise that
$$\left| \frac{2n - 16}{n^2 + 5} \right| \leq \frac{2n+16}{n^2}$$
I don't find it obvious, con you find the operation he made to conclude that?

You can use the double inequality

$$0 < \frac{\sqrt{1+n}}{n} <\frac{2}{\sqrt{1+n}}$$, valid for n>1.
Hi dextercioby,
Thanks! Sounds trivial now, but how did you get that insight?

Deveno
Hi dextercioby,
Thanks! Sounds trivial now, but how did you get that insight?
if n > 1, then n+1 < n+n = 2n.

but n+1 = (√(n+1))(√(n+1)), for positive n.

(this is a standard trick with things involving square roots:

1/√x = √x/x. in this particular instance we use the reciprocals:

√x = x/√x. so:

√(n+1)/n = (n+1)/[n√(n+1)] < 2n/[n√(n+1)] = 2/√(n+1).

the only real insight, besides the "algebraic wizardry"

is that n > 1 implies n+1 < 2n).

SammyS
Staff Emeritus
Homework Helper
Gold Member
...

I have a related question: my teacher used in part of an exercise that
$$\left| \frac{2n - 16}{n^2 + 5} \right| \leq \frac{2n+16}{n^2}$$
I don't find it obvious, con you find the operation he made to conclude that?
$\displaystyle \left| {2n - 16} \right| \leq {2n+16}$ by using the triangle inequality.

$\displaystyle n^2 + 5 \ge n^2>0$ pretty obviously.

Therefore: $\displaystyle \frac{1}{n^2 + 5}\le\frac{1}{n^2}$