I've got the following question: State the value of $a$ that makes the function continuous, where $$f(x) = \begin{cases} x^2 + 3x + 5, && \text{when $x > 1$}\\ a, && \text{when $x = 1$}\\ 12x - 3, && \text{when $x < 1$.} \end{cases} $$
I've found the following limits by substitution: \begin{align*} \lim_{x \to 1+} x^2 + 3x + 5 & = (1)^2 + 3(1) + 5 = 9\\ \lim_{x \to 1-} 12x-3 & = 12(1) -3 = 9 \end{align*}
Would this mean that in order to make the function continuous, $a$ is simply $= 9$?
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$\begingroup$Indeed, $a=9$ is the answer. You really have three functions under consideration, the two that you do (call them $f$ and $h$ respectively) and $g(x)=a$. Your function will be continuous when
$$\lim_{x\to 1^-}f(x)=g(1)=\lim_{x\to 1^+}h(x)$$
I think you're being put off by how easy it was to choose $a=9$, so I'll make the problem a little more complicated and solve that in hopes that it is illuminating.
If we make the function a little more complicated, such as letting $g(x)=x^3+7a^2x+1$, then we have to solve for $a$, obtaining
\begin{align*} 9&=1+7a^2\cdot1+1\\ 9&=7a^2+2\\ 7&=7a^2\\ a&=\pm 1 \end{align*}
We get $9=1+7a^2\cdot 1+1$ simply by declaring that $g(1)=9$, as the above equation requires. Notice that here there are two solutions. In general examples, there can be infinitely many potential values of $a$ that work (say if $g$ involves a sine function), though for a polynomial there will always be finitely many solutions. The fact that there are two comes from the fact that the $a$ term is squared. Recall that a polynomial of degree two has up to two distinct solutions, and the equation $7=1+7a^2+1$ is a polynomial, with $a$ as the variable.
What you did is the same thing, except it's weirdly simple since you have $g(x)=a$, and so $a=9$ occurs immediately.
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