<s>[INST] Multiplying both sides of an equation when there's a limit on one side? 
Determine the  value of $a\in\mathbb{R}$, such that $\displaystyle\lim_{x\to 1}\dfrac{x^2+(3-a)x+3a}{x-1}=7$

My attempt:
\begin{align*}
&\lim_{x\to 1} \dfrac{x^2+(3-a)x+3a}{x-1}=7\\
&\implies\lim_{x\to 1} x^2+(3-a)x+3a=7x-7\\
&\implies\lim_{x\to 1} x^2-4x-a(x-3)=-7\\
&\implies1^2-4(1)-a(1-3)=-7\\
&\implies a=-2
\end{align*}
My question is can I multiply both sides of my first line by $x-1$, even though the equation has a $\lim_{x\to 1}$ on the left side, as opposed to a normal equation?
 [/INST] There is a safer way to think. If you try to plug $1$ into the limit, you'll get a division by zero. So, your only hope for this limit to make sense, is if $x^2 + (3 - a)x + 3a$ is divisible by $x - 1$, so you can simplify and eliminate the division by zero. This means that $1$ must be a root of $x^2 + (3-a)x + 3a$. So: $$1 + (3-a) + 3a = 0 \implies 4 +2a = 0 \implies a = -2.$$
Fortunately, we indeed have: $$\lim_{x \to 1} \frac{x^2 + 5x - 6}{x - 1} = \lim_{x \to 1}\frac{(x+6)(x-1)}{x-1 } = \lim_{x \to 1} x + 6 = 7.$$
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