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If I have a function $f$, and the limit exists at $a$, and $\lim\limits_{x \to a} f(x) = \infty $ or $\lim\limits_{x \to a} f(x) = k $, where $k$ is a constant, does that mean that the function is continuous on some interval in the domain of $f$?

I think this is true, but I am not sure. Because for example, if $\lim\limits_{x \to a} f(x) = \infty $, if I graph a function that fulfills this, there is at least some interval in its domain where it is continuous.

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    $\begingroup$ Depends on what $f(a)$ is. The limit existing is not enough for continuity. (You can generalize functions to allow $f(a)=\infty,$ but that usually isn’t allow. For real-valued functions, there is no way for $f$ to be continuous at $a$ if the limit doesn’t exist or is $\pm\infty.$ But if the limit is $k,$ it is continuous iff $k=f(a).$) $\endgroup$ Oct 24 at 2:08
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The existence of a limit does not imply that the function is continuous somewhere. Some counterexamples: $$\text{Let }f_1(x) = \begin{cases} 0&x=0\\ \frac1{x^2}&x\in\mathbb{Q}\backslash\{0\}\\ \frac1{2x^2}&x\notin\mathbb{Q} \end{cases}$$

$$\text{and let }f_2(x) = \begin{cases} 1&x=0\\ x&x\in\mathbb{Q}\backslash\{0\}\\ -x&x\notin\mathbb{Q}\end{cases}$$ Here, we can see that $\lim\limits_{x\to0}f_1(x)=\infty$ and $\lim\limits_{x\to0}f_2(x)=0$, but $f_1$ and $f_2$ are nowhere continuous.

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Let $$f(x) = \begin{cases} x \quad{} &\text{if $x \ne 0$ and $x$ is rational}\\ 0 &\text{if $x$ is irrational} \\ 1 & \text{if $x = 0$}\end{cases}.$$ You can see that $\lim_{x \to 0} f(x) = 0$, but $f$ is continuous nowhere.

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