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?
 A: 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
$$\lim_{x\to 0}f(g(x))=4\ne 3$$
A: Suppose $\alpha\notin g(\mathbb R\setminus\{\beta\})$.
Let $\epsilon>0$ be arbitrary. Then there exists a $\delta>0$ such that $0<|x-\alpha|<\delta\Rightarrow|f(x)-L|<\epsilon$.
Furthermore, there exists a $\eta>0$ such that $0<|y-\beta|<\eta\Rightarrow|g(y)-\alpha|<\delta$. Since $g(y)\ne\alpha$ by our hypothesis, we have $0<|g(y)-\alpha|<\delta$. Then, by my statement above, this implies that $|f(g(y))-L|<\epsilon$, which shows that $\lim_{y\to\beta}f(g(y))=L$.
P.S.: You can get away with assuming that for a small enough $\epsilon>0$ there are no $x\in\mathbb R\setminus\{\beta\}$ with $|x-\beta|<\epsilon$ 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.
A: This is true, and the $\varepsilon-\delta$-approach isn't that tedious, actually: For every $\varepsilon>0$ there is a $\lambda>0$ such that $\vert f(x)-L\vert<\varepsilon$ for all $x$ with $\vert x-\alpha\vert<\lambda$. Also, for each $\lambda>0$, there is a $\delta>0$ such that $\vert g(y)-\alpha\vert<\lambda$ for all $y$ with $\vert y-\beta\vert<\delta$. Put both facts together to get that for all $\varepsilon>0$ there is a $\delta>0$ such that $\vert f(g(y))-L\vert<\varepsilon$ for all $y$ with $\vert y-\beta\vert<\delta$.
Edit: To be clear, this assumes the following definition of the limit of a function. Let $X\subseteq\mathbb R$, $f:X\to \mathbb R$, $x_0\in X$. Then $\lim_{x\to x_0}f(x)=y$ if for all $\varepsilon>0$ there exists a $\delta>0$ such that for all $x\in X$ with $\vert x-x_0\vert<\delta$ we have $\vert f(x)-y\vert<\varepsilon$. In a more topological language, for every open neighborhood $V$ of $y$ in $X$ there is an open neighborhood $U$ of $x_0$ such that $f(U)\subseteq V$. This is an contrast to a slightly different definition where the condition need only hold for all $x\in X\backslash\{x_0\}$, or in a topological language, for all punctured neighborhoods of $x_0$. The two are not equivalent, and my argument only holds with the former. To work with the latter, $g$ must also be continuous.
