<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] $$\lim_{x\to 1}\dfrac{x^2+(3-a)x+3a}{x-1}=7$$
$$\lim_{x\to 1}\dfrac{x^2+(3-a)x+3a}{x-1}=\lim_{x\to 1}7$$
$$\lim_{x\to 1}\dfrac{x^2+(3-a)x+3a}{x-1}-\lim_{x\to 1}7=0$$
$$\lim_{x\to 1}\left(\dfrac{x^2+(3-a)x+3a}{x-1}-7\right)=0$$
$$\lim_{x\to 1}\dfrac{x^2+(3-a)x+3a-7x+7}{x-1}=0$$
$$\lim_{x\to 1}\dfrac{x^2+(-4-a)x+3a+7}{x-1}=0$$
To the limit exists and be $0$ the roots of the numerator must be one with multiplicity two (one to eliminate the $x-1$ and the other to make the limit zero). Therefore:
$$-4-a=-(1+1)$$
$$3a+7=1\cdot 1$$
Both equations have a unique solution $a=-2$
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