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A limit identity that blows my mind is the following:

$\lim\limits_{n\rightarrow \infty}x^{1/n} = 1$

for $( x > 0) $

Rather than just accepting this limit as a fact, I am wondering whether there is a nice way to prove it (preferably visually).

A huge chunk of my confusion comes from the fact that raising a number to a non whole power ex: $3^{0.35}$ can only be computed through computers.

In a similar manner square rooting a number that is not a perfect square, perfect cube, etc... can only be done through computers! Example: Let x = 3 and N = 5 thus, $3^{1/5}$.

Assuming that I cannot fully visualize this, how can I go on to say something like $0.35^{1/250} \approx 1$

Edit: Thanks to everyone who helped solve this. With the exception of the guy who claimed "anything raised to zero = one" , I loved all the proofs and comments.

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    $\begingroup$ You can certainly compute $a=3^{35/100}$ by hand, if you can compute integer powers. For example, since $1^{100} < 3^{35}$ and $2^{100} > 3^{35}$, $a$ must be $1.$xxxx and not $2.$xxxx. Similarly, since $1.4^{100} < 3^{35}$ and $1.5^{100}>3^{35}$, $a$ must be of the form $1.4$xxx. Continuing, you can get as many decimals you want. $\endgroup$
    – JWL
    Sep 11 '19 at 3:37
  • $\begingroup$ If $x>0$ then for every $x$ we have that $$\ln(x^\frac{1}{n})=\frac{1}{n}\ln (x)\to 0 ~~\text{as}~~n\to\infty$$ Then, as $f(u)=e^u$ is continuous at $0$, we have $$\lim_{n\to\infty}x^\frac{1}{n}=\lim_{n\to\infty}e^{\frac{1}{n}\ln x}=\lim_{u\to 0}e^u=e^0=1$$ $\endgroup$
    – Axion004
    Sep 11 '19 at 5:10
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Suppose that the limit of the function is some value $y$. Then, $\lim_{n\rightarrow\infty} x^{1/n} = y$. Take the natural log of both sides. Since the natural log is continuous for positive values, you can pass it through the limit and get $$ \lim_{n\rightarrow\infty} \ln{x^{1/n}} = \lim_{n\rightarrow\infty} \frac{\ln{x}}{n} = \ln y.$$

Then, as $n \rightarrow \infty$ you get that $$0 = \ln y.$$ Then, $$e^0 = e^{\ln y} \rightarrow 1 = y.$$

Therefore, $\lim_{n\rightarrow\infty} \ln x^{1/n} = 1$.

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  • $\begingroup$ Thank you for this proof. $\endgroup$
    – Josue
    Sep 11 '19 at 3:20
  • $\begingroup$ NB this proof only works if you assume that $\lim_{n \to \infty} x^{1/n} $ exists. This doesn't pose that much of a difficulty, though, since $x^{1/n}$ is either increasing and bounded above by $1$ ($x \leq 1$) or decreasing and bounded below by $1$ ($x \geq 1$). In both cases, we may conclude that the limit exists. $\endgroup$ Sep 11 '19 at 4:44
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First observe that for any $a > 0$, $x^{1 \over n} < a$ is equivalent to $x < a^n$.

Suppose $a > 1$. Then if $n$ is large enough, $x < a^n$ will be satisfied since $a > 1$. So $x^{1 \over n} < a$.

Similarly, if $b < 1$, if $n$ is large enough, $x > b^n$ will be satisfied since $b < 1$. So $x^{1 \over n} > b$.

Letting $b = 1 - \epsilon$ and $a = 1 + \epsilon$ for small $\epsilon > 0$, we see that for $n$ large enough we have $$1 - \epsilon < x^{1 \over n} < 1 + \epsilon$$ Equivalently, for $n$ large enough we have $$|x^{1 \over n} - 1| < \epsilon$$ This is exactly the statement that $\lim_{n \rightarrow \infty} x^{1 \over n} = 1$.

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  • $\begingroup$ For OP's benefit: to see why we can find $n$ large enough that $x < a^n$ when $a > 1$, observe that $a^n = (1 + (a-1))^n \geq 1 + n(a- 1)$ . The inequality follows from the binomial theorem and the fact that $1$ and $a-1$ are both nonnegative. The quantity $1 + n(a-1)$ can be made as large as we like by the Archimedean property of the real numbers. Thus $a^n$ can be made as large as we like. $\endgroup$ Sep 11 '19 at 4:10
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We examine three cases; $(1)$ $x=1$, $(2)$ $x>1$, $(3)$ $x<1$.


Case $\displaystyle (1)$: $\displaystyle x=1$.

When $x=1$, $x^{1/n}=1$ for all $n\in \mathbb{N}$ and the result is trivial.


Case $\displaystyle (2)$: $\displaystyle x>1$.

If $x>1$, then for any $n\in \mathbb{N}$, $x^{1/n}>1$. We define the sequence $y_n$ by the expression

$$y_n\equiv x^{1/n}-1\tag 1$$

where we see that $y_n>0$. Rearranging $(1)$, we write $x=(1+y_n)^n$ whence from Bernoulli's Inequality we find that

$$x>1+ny_n\tag2$$

Solving $(2)$ for $y_n$ reveals

$$0<y_n<\frac{x-1}{n}\tag3$$

Application of the squeeze theorem to $(3)$ yields

$$\lim_{n\to\infty}y_n=0$$

from which we find from $(1)$

$$\lim_{n\to\infty}x^{1/n}=1$$


Case $\displaystyle (3)$: $\displaystyle x<1$.

If $x<1$, then for any $n\in \mathbb{N}$, $x^{1/n}<1$. We define the sequence $z_n$ by the expression

$$z_n\equiv \frac1{x^{1/n}}-1\tag 4$$

where we see that $z_n>0$. Rearranging $(4)$, we write $x=\frac1{(1+z_n)^n}$ whence from Bernoulli's Inequality we find that

$$x<\frac1{1+nz_n}\tag5$$

Solving $(5)$ for $z_n$ reveals

$$0<z_n<\frac{1/x-1}{n}\tag6$$

Application of the squeeze theorem to $(6)$ yields

$$\lim_{n\to\infty}z_n=0$$

from which we find from $(4)$

$$\lim_{n\to\infty}x^{1/n}=1$$


And we are done!

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  • $\begingroup$ Please let me know how I can improve my answer. I really want to give you the best answer I can. $\endgroup$
    – Mark Viola
    Oct 12 '19 at 17:16
  • $\begingroup$ And please feel free to accept and up vote an answer as you see fit. $\endgroup$
    – Mark Viola
    Nov 11 '19 at 23:30
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Computers

It is not true that "raising a number to a non whole power," like $3^{0.35}$, "can only be computed through computers". Anything that a computer can do can be done with manual computation. One way is by using logarithms and exponentials, tables of which were computed long before computers. Computers just make the computations faster and generally less error prone.


The Limit

Bernoulli's Inequality says that for $n\ge1$ and $x\gt-1$, $$ 1+x\le\left(1+\frac xn\right)^n\tag1 $$ raising to the $1/n$ power $$ (1+x)^{1/n}\le1+\frac xn\tag2 $$ Furthermore, for $n\ge1$ and $x\gt-1$, so that $\frac{x}{1+x}\lt1$, Bernoulli's Inequality says that $$ 1-\frac{x}{1+x}\le\left(1-\frac{x}{n(1+x)}\right)^n\tag3 $$ raising to the $-1/n$ power $$ \begin{align} (1+x)^{1/n} &\ge1+\frac{x}{n(1+x)-x}\\ &\ge1+\frac{x}{n(1+x)}\tag4 \end{align} $$ Thus, for $n\ge1$ and $x\gt-1$, $$ 1+\frac{x}{n(1+x)}\le(1+x)^{1/n}\le1+\frac xn\tag5 $$ The Squeeze Theorem applied to $(5)$ says that for $x\gt-1$, $$ \lim_{n\to\infty}(1+x)^{1/n}=1\tag6 $$ then substituting $x\mapsto x-1$ says that for $x\gt0$, $$ \lim_{n\to\infty}x^{1/n}=1\tag7 $$

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