Question

Suppose \(\lim_{x\rightarrow a}f(x) = c\). How can I show that \(\lim_{x\rightarrow a} \sqrt[n]{f(x)} \) exists for all n in \(\mathbb{N}\)?

Answer 1

looks like an inductive proof

Answer 2

that doesn't mean its bad :)

Answer 3

with n=1 , thats true by limit rules

Answer 4

I guess we are suppose to just assume a and c>0?

Answer 5

oh, when n=1 , its not square rooting at all !

Answer 6

\(a \in \mathbb{R}\) and c>0

Answer 7

when n=1, you have lim (f(x)) ^(1/1) = lim f(x)

Answer 8

A side question- is the limit c^(1/n)?

Answer 9

yes

Answer 10

because

Answer 11

draw button died :/

Answer 12

lim (sqrt(f(x)) = sqrt( lim(f(x))

Answer 13

lim (sqrt(f(x)) = sqrt( lim(f(x)) <- prove it!

Answer 14

actually that is the case n=2

Answer 15

the statement is more general

Answer 16

lim [ f(x)^(1/n)] = ( lim(f(x)) ^(1/n)

Answer 17

I'll try induction. Have to go for a while, will get back to this soon.

Answer 18

lim [ f(x) ^(1n)] = [ lim(f(x) ] ^(1/n)

Answer 19

freckles wanted to say something :)

Answer 20

say we have
\[f(x)=\frac{c}{a} x \\ \lim_{x \rightarrow a} f(x)=c \\ \lim_{x \rightarrow a}\sqrt{f(x)}=\lim_{x \rightarrow a} \sqrt{ \frac{c x}{a}}= \sqrt{c}\]
But I think a can be a real number not equal to 0 for this function
since we cannot divide by 0
-
i'm just worried about the restrictions is all

Answer 21

this question has multiple levels of generality ( i dont know a better word for it)

Answer 22

you *found* a special case of n = 2 :)

Answer 23

it really stinks you can't edit your comments

Answer 24

i know what you meant though

Answer 25

And I was just picking a function that satisfied c>0 and a is a real number
but I think there needs to be more restrictions on a or maybe on f

Answer 26

Maybe I'm being difficult

Answer 27

yeah i see what you mean

Answer 28

so you would just have to say, a =/= 0

Answer 29

We shall prove it in a more general form: \(\lim_{x\rightarrow a} \sqrt[n]{f(x)^m}\).

First, we'll show that if \(\lim_{x\rightarrow a}f(x) = c\), then \(\lim_{x\rightarrow a}f(x)^m \) exists and \(\lim_{x\rightarrow a}f(x)^m =(\lim_{x\rightarrow a}f(x))^m = c^m\). This shall be proved by induction.

When m=2,
\(\lim_{x\rightarrow a}f(x)^2 = \lim_{x\rightarrow a}f(x) \cdot \lim_{x\rightarrow a}f(x) =(\lim_{x\rightarrow a}f(x))^2= c^2\)

Suppose \(\lim_{x\rightarrow a}f(x)^k = (\lim_{x\rightarrow a}f(x))^k=c^k\) for some positive integers k.

Consider m=k+1,
\(\lim_{x\rightarrow a}f(x)^{k+1} \)
\(= \lim_{x\rightarrow a}f(x)^k \cdot \lim_{x\rightarrow a}f(x) \)
\(= (\lim_{x\rightarrow a}f(x))^k \cdot\lim_{x\rightarrow a}f(x) \)
\(= c^k\cdot c = c^{k+1}\)
Hence, by principle of mathematical induction, if \(\lim_{x\rightarrow a}f(x) = c\), \(\lim_{x\rightarrow a}f(x)^m = (\lim_{x\rightarrow a}f(x))^m =c^m\) for all positive integers m.

Now, let \(g(x) = \sqrt[n]{f(x)^m}\), then we have \(g(x)^n = f(x)^m\), and \(\lim_{x\rightarrow a}g(x)^n = \lim_{x\rightarrow a}f(x)^m\)
From the above, we've shown that \(\lim_{x\rightarrow a}f(x)^m = c^m\).
Hence, we have \((\lim_{x\rightarrow a}g(x))^n = \lim_{x\rightarrow a}g(x)^n = \lim_{x\rightarrow a}f(x)^m = c^m\)
That is,
\((\lim_{x\rightarrow a}g(x))^n= c^m\)Taking the nth root on both sides, we have
\(\lim_{x\rightarrow a}g(x)= c^\frac{m}{n}\)
Hence, we've proved that if \(\lim_{x\rightarrow a}f(x) = c\), \(\lim_{x\rightarrow a} \sqrt[n]{f(x)^m}\) exists, and \(\lim_{x\rightarrow a} \sqrt[n]{f(x)^m}= c^\frac{m}{n}\) for all \(m,n\in\mathbb{N}\)

Let m=1, we have
if \(\lim_{x\rightarrow a}f(x) = c\), then \(\lim_{x\rightarrow a} \sqrt[n]{f(x)}\) exsits, and \(\lim_{x\rightarrow a} \sqrt[n]{f(x)}= c^\frac{1}{n}\) for all \(n\in\mathbb{N}\)