# Math Help - question regarding limit to infinity

1. ## question regarding limit to infinity

Hi my question is this:
If two functions, say f and g are defined on an open set $(a,\infty)$, $a\in\mathbb{R}$ where
$\lim_{x \to \infty}f(x)=L$ and $\lim_{x \to \infty}g(x)=+\infty$,
how can I show that $\lim_{x\to\infty}(f\circ g)(x)=L$

2. Originally Posted by dannyboycurtis
Hi my question is this:
If two functions, say f and g are defined on an open set $(a,\infty)$, $a\in\mathbb{R}$ where
$\lim_{x \to \infty}f(x)=L$ and $\lim_{x \to \infty}g(x)=+\infty$,
how can I show that $\lim_{x\to\infty}(f\circ g)(x)=L$
Since $g(x)\to\infty$ as $x\to\infty$, then $\lim_{x\to\infty}f(g(x)) = \lim_{g(x)\to\infty}f(g(x))$, which is $L$ by definition.

(See tonio's post below for a full-blown $\epsilon$-proof.)

3. Originally Posted by dannyboycurtis
Hi my question is this:
If two functions, say f and g are defined on an open set $(a,\infty)$, $a\in\mathbb{R}$ where
$\lim_{x \to \infty}f(x)=L$ and $\lim_{x \to \infty}g(x)=+\infty$,
how can I show that $\lim_{x\to\infty}(f\circ g)(x)=L$

Choose any e > 0 ==> there exists a real number M s.t. |f(x) - L| < e whenever x > R.

Since g(x) --> oo when x --> oo there exists a number T s.t. g(x) > R whenever x > T ==> for ANY x > T you get

|f(g(x)) - L| < e , and this means f(g(x)) --> L when x --> oo

Tonio

4. Originally Posted by tonio
Choose any e > 0 ==> there exists a real number M s.t. |f(x) - L| < e whenever x > R.
You mean there exists a real number R s.t...

5. Or he means "whenever x> M".