# 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 $\lim_{x\to 0^+} f\left( \frac{1}{x} \right) = L$.

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

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

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

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

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