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This question arises from one of the Berkeley Problems in Mathematics, fall 1981. It asks me to find a counterexample to the following claim:

Suppose $\lim\limits_{x\rightarrow x_{0}}f(x)=a$, $\lim\limits_{t\rightarrow a}g(t)=b$, prove or find a counterexample for the claim $\lim\limits_{x\rightarrow x_{0}}g(f(x))=b$.

I feel this should not be true, for we can select functions with infinite 'singularities' near the origin like $x\sin[1/x]$. Then for each value $x=\frac{2}{k\pi}$ we should have the value to be equal to $x$. But this is not a counterexample. Looking up the book the authors give the following example:

$f=g=1$ when $t\not=0$, $f=g=0$ when $t=0$.

The authors then claimed that $\lim_{x\rightarrow 0}g(f(x))=1$ while $\lim_{t\rightarrow 0}g=0$. Unless I am confused with the definition of the limit, that there exists $\lambda$ such that for any $\epsilon$, if $|x-x_{0}|\le \lambda$, then $f(x)-A|\le \epsilon$, I do not see why $\lim_{t\rightarrow 0}g=0$ since $t=0$ is a removable singularity for $g$. So assuming I am wrong somewhere, I want to ask:

1): why the limit of $g$ at 0 is 0?

2): is there better counterexamples?

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Actually, it looks like the authors define $$f(t) = g(t) = \begin{cases} 0 \text{ for } t \neq 0 \\ 1 \text{ for } t = 0, \end{cases}$$ and then state that $\lim_{t \to 0} f(t) = \lim_{t \to 0} g(t) = 0$, and $\lim_{t \to 0}f(g(t)) = 1$. – Jesse Madnick Jul 22 '12 at 5:52
yeah, I think I copied it wrong. I see. – Bombyx mori Jul 22 '12 at 5:56
up vote 3 down vote accepted

I think you just made a careless error. In the example provided, the limit of $g$ at 0 is not 0; it is 1, as you said. Here $x_0 = 0$, so that $a = 1$. So lim$_{t\rightarrow a}{g(t)}$ = lim$_{t\rightarrow 1}{g(t)} = 0$. And lim$_{x\rightarrow x_0}{g(f(x))} = $ lim$_{x\rightarrow 0}{g(f(x))} = 1$. This is probably as simple a counterexample as one could hope for.

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