Let $f$ and $g$ be defined on $(a,\infty)$ and suppose $\displaystyle\lim_{x\rightarrow\infty}(f)=L$ and $\displaystyle\lim_{x\rightarrow\infty}(g)=\infty$. Prove that $\displaystyle\lim_{x\rightarrow\infty}(f\circ g)=L$.

I am having trouble proving this using $\epsilon-\delta$ definition. Any alternatives? Or how can it be proved using the epsilon-delta definition?

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    $\begingroup$ Do you know the definition? $\endgroup$ – Git Gud Oct 20 '13 at 15:53

We know that $\lim_{x \to \infty}f(x) = L$ by definition means that for any $\varepsilon > 0$ there exists an $N$, such that $$x > N\implies |f(x) -L|<\varepsilon, \tag{1}$$ and also that since $\lim_{x \to \infty}g(x) = \infty$, by definition for any $O$ there exists an $M$ such that $$x > M \implies g(x)>O. \tag{2}$$

Now we need to show using these definitions that $\lim_{x \to \infty}f(g(x)) = L$. This just means showing that for any $\varepsilon>0$ there exists a $P$ such that $$x>P\implies|f(g(x))-L|<\varepsilon. \tag{3}$$

Since we have the liberty to choose any $O$ we want (the definition guarantees that for any $O$ an $M$ exists), just choose $O = N$. Then $(2)$ tells us that there exists some $M$ such that $x > M \implies g(x) > N$. But then our definition of the limit as $x$ approaches infinity of $f$ tells us that $g(x) > N \implies |f(g(x)) - L| < \varepsilon.$ And now we're done! For any $\varepsilon > 0$ there is an $M$ such that $$x>M\implies g(x) > N \implies |f(g(x))-L|<\varepsilon.$$ I've added the middle step $(g(x) > N)$ for clarity, but if you take it out then you can clearly see that this chain of statements is just a restatement of $(3)$, so our proof is complete. $\square$


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