# Finding the continuity of function

A function f is defined by: $$f(x) = \begin{cases} x^4 + 1, \ \text{ if }x < 0,\\ \\ 0, \ \ \ \ \ \ \ \ \ \text{ if } x = 0,\\ \\ x^2 + 1, \ \text{ if } x > 0. \end{cases}$$

Find the limit as $x \to 0$

I know that the one-sided limits as $x\to 0^+$ and $x\to 0^-$ are both $1.$ However, the $f(x) = 0$ part confuses me a bit. My guess is that the limit does not exist, my reasoning being that because

$\lim_{x\to a} c = c$ , then

$\lim_{x \to 0} 0 = 0$

If this is not correct, then what bit of fundamental understanding am I missing?

The limit as x approaches zero exists and is $1$ as you say. The limit does not care about what the function value is exactly at zero. The fact that $f(0)=0 \neq \lim_{x \to 0} f(x)$ says the function is not continuous there.