Question about finding the limit of a^(1/n)

[Clara Chung]

Homework Statement

First part (a>1) of the proof:
Denote h = a^(1/n) - 1> -1
Then a = (1+h)^(n) >= 1+nh
so h <= (a-1) / n
Assume a > 1, so that 0 <h <= (a-1)/h n tends to infinity
By sandwich principle, lim n tends to infinity of h is 0

Homework Equations

The Attempt at a Solution

Why is h > 0 when a > 1? Did the proof assume that a^(1/n) tends to one, so h = a^(1/n) - 1 > 0?

 

[BvU]

Hi,

Let me guess: this is an example from textbook or blackboard (c.q. its modern equivalent) and it leaves you wondering ?
And the frst part (or the whole exercise?) is to prove this for $a>1$ -- for which we know that $a^{\rm any\ power} > 0$ ?

Did the proof assume that a^(1/n) tends to one

That would invalidate the proof: you can't prove something by using what you want to prove as a correct assumption.

If my guess above is correct, the author expects that $\displaystyle \lim_{n\rightarrow\infty} a^{({1\over n})} = 1$ and embarks on an expedition to prove that $\displaystyle \lim_{n\rightarrow\infty} a^{({1\over n})} - 1 = 0$

Why is h > 0 when a > 1?

Mysterious. All I see so far is that $h > -1$

I also have difficulty understanding that $(1+h)^n \ge 1+nh$, can you explain how that works for $h>-1$ ?

 

[Clara Chung]

Hi,

Let me guess: this is an example from textbook or blackboard (c.q. its modern equivalent) and it leaves you wondering ?
And the frst part (or the whole exercise?) is to prove this for $a>1$ -- for which we know that $a^{\rm any\ power} > 0$ ?

That would invalidate the proof: you can't prove something by using what you want to prove as a correct assumption.

If my guess above is correct, the author expects that $\displaystyle \lim_{n\rightarrow\infty} a^{({1\over n})} = 1$ and embarks on an expedition to prove that $\displaystyle \lim_{n\rightarrow\infty} a^{({1\over n})} - 1 = 0$

Mysterious. All I see so far is that $h > -1$

I also have difficulty understanding that $(1+h)^n \ge 1+nh$, can you explain how that works for $h>-1$ ?

How do you see that h > -1 only?
If h > -1, we can use the bernoulli's inequality, so (1+h)^(n) >= 1+nh ....

 

[BvU]

If h > -1, we can use the bernoulli's inequality, so (1+h)^(n) >= 1+nh ....

Very good - that's what I was fishing for.

How do you see that h > -1 only?

It's what you wrote

Denote h = a^(1/n) - 1> -1

and it can only be correct if $a^{(1/n)}>0$ -- fortunately, that is the case for $a>1$.

What's stilll missing is a proof that $h>0$ -- or am I mistaken ?

 

[Clara Chung]

Very good - that's what I was fishing for.
It's what you wrote
and it can only be correct if $a^{(1/n)}>0$ -- fortunately, that is the case for $a>1$.

What's stilll missing is a proof that $h>0$ -- or am I mistaken ?

The teacher wanted to prove that lim a^(1/n) equals to 1, but he used the information that h>0 in order to use the sandwich theorem. is the proof invalid in this case?

 

[BvU]

Invalid is a strong term... but I would like to know how $\ h\ge-1 \$ can come back a few lines later as $\ \ 0<h \$...