Does this property of limits hold when we compose functions? Suppose we have two functions $f,g:\mathbb{R}\to{\mathbb{R}}$ such that:
$$\lim_{x\to{}\alpha}{f(x)}=L \text{ and }\lim_{y\to{\beta}}{g(y)}=\alpha.$$
Is then true that:
$$\lim_{y\to{}\beta}{f(g(y))=L}?$$
My intuition tells me yes. Using an $\epsilon-\delta$ approach seems very long and tedious. Is that the case? Or is there a nice way of showing this to be true. Or, is there a nice counterexample?
Here is a counterexample :
$$g(x)=1\;if\; x\ne 0\; and\;g(0)=2$$
$$f(x)=3\;if\;x\ne 1\; and\;f(1)=4$$
thus
$$f(g(x))=4\;if\;x\ne 0\;and\;f(g(0))=3$$
take $$\alpha=1\;and \;\beta=0$$
we have
$$\lim_{x\to 1}f(x)=3=L$$
$$\lim_{x\to 0}g(x)=1$$
but
$$\lim_{x\to 0}f(g(x))=4\ne 3$$
Suppose $\alpha\notin g(\mathbb R\setminus\{\beta\})$.
Let $\epsilon&gt;0$ be arbitrary. Then there exists a $\delta&gt;0$ such that $0&lt;|x-\alpha|&lt;\delta\Rightarrow|f(x)-L|&lt;\epsilon$.
Furthermore, there exists a $\eta&gt;0$ such that $0&lt;|y-\beta|&lt;\eta\Rightarrow|g(y)-\alpha|&lt;\delta$. Since $g(y)\ne\alpha$ by our hypothesis, we have $0&lt;|g(y)-\alpha|&lt;\delta$. Then, by my statement above, this implies that $|f(g(y))-L|&lt;\epsilon$, which shows that $\lim_{y\to\beta}f(g(y))=L$.
P.S.: You can get away with assuming that for a small enough $\epsilon&gt;0$ there are no $x\in\mathbb R\setminus\{\beta\}$ with $|x-\beta|&lt;\epsilon$ such that $\alpha=g(x)$.
When this is not the case, there is a sequence of points $y_n\in\mathbb R\setminus\{\beta\}$ with $y_n\to\beta \ (n\to\infty)$ such that $g(y_n)=\alpha$. Then, $\lim_{n\to\infty}f(g(y_n))=f(\alpha)$, so that $\lim_{y\to\beta}f(g(y))=f(\alpha)$ if this limit exists.