# Thread: proof of 2 limits

1. ## proof of 2 limits

hi im stuck with this question for a while. its a real analysis topic... i was wondering if anyone can help me by
the question:
f is defined on (0,infinity) to R.

i want to prove that lim x->infinity f(x) =L if and only if lim x->0 f(1/x)=L ..the x->0 is a positive so its right limit.

my solution so far

i decided to prove each one seperately and let y =f(x),
but the thing is i was getting confused with the definitions and what to do next when using the definitions.

so for the first equation

lim x->infinity f(x) =L

let epsilon>0, N is in natural numbers, y>=N => ! y- L ! < epsilon

this doesnt seem quite right and i wasnt to sure what to do next.

so for the second limit

lim x->0 f(1/x)=L ..the x->0 is a positive so its right limit.

let epsilon>0

tilda = 1/epsilon

next 0<x- 0 <1/epsilon

this gives ! f(1/x) - L ! < epsilon

since y= f(x) where the x has to be replaced then ! f(x)- L!< epsilon ...hence the quesiton is proved...

can some one check this and help me correct it please thanks

2. Originally Posted by dopi
hi im stuck with this question for a while. its a real analysis topic... i was wondering if anyone can help me by
the question:
f is defined on (0,infinity) to R.

i want to prove that lim x->infinity f(x) =L if and only if lim x->0 f(1/x)=L ..the x->0 is a positive so its right limit.
You are missing something, it should be $\displaystyle \lim_{x\to 0^+} f\left( \frac{1}{x} \right) = L$.

Let $\displaystyle f0,+\infty)\mapsto \mathbb{R}$ and define $\displaystyle g0,+\infty)\mapsto \mathbb{R}$ as $\displaystyle g(x) = f\left( \frac{1}{x} \right)$.

I will use the sequencial approach to limit because this approach is the nicest. Let $\displaystyle x_n$ be a sequence in $\displaystyle (0,\infty)$ converging to $\displaystyle 0$, i.e. $\displaystyle \lim \ x_n = 0$. Then, we know (this is an easy theorem, but try to prove it) $\displaystyle \lim \ \frac{1}{x_n} = +\infty$. Thus, $\displaystyle \frac{1}{x_n}$ is a sequence in $\displaystyle (0,\infty)$ approaching $\displaystyle +\infty$. Thus, this means $\displaystyle \lim f\left( \frac{1}{x_n} \right) = L$. This implies then that $\displaystyle \lim g(x_n) = \lim f\left( \frac{1}{x_n} \right) = L$. Q.E.D.

The theorem I used was let $\displaystyle x_n$ be a sequence of positive real numbers which is convergent to zero. Then $\displaystyle \lim \frac{1}{x_n} = +\infty$. This theorem goes both ways.

3. Given $\displaystyle \lim _{x \to \infty } f(x) = L$ means that:
$\displaystyle \left( {\varepsilon > 0} \right)\left( {\exists B > 0} \right)\left[ {x > B \Rightarrow \left| {f(x) - L} \right| < \varepsilon } \right]$.

Select $\displaystyle \left( {\delta > 0} \right)\left[ {\delta < \frac{1}{B}} \right]$. Now if $\displaystyle 0 < x < \delta \Rightarrow B < \frac{1}{x} \Rightarrow \left| {f\left( {\frac{1}{x}} \right) - L} \right| < \varepsilon$.
Or $\displaystyle \lim _{x \to 0^ + } f\left( {\frac{1}{x}} \right) = L$.