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?
 A: 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: $$\lim_{x \to 1} \frac{x^2 + 5x - 6}{x - 1} = \lim_{x \to 1}\frac{(x+6)(x-1)}{x-1 } = \lim_{x \to 1} x + 6 = 7.$$
A: No, you can't multiply both sides by $x - 1$, or at least not the way you think. In the limit as written, $x$ only exists within the limit. If you multiply both sides by something, that something is outside the limit:
$$(\text{something})\biggl(\lim_{x\to 1}\frac{x^2+(3-a)x+3a}{x-1}\biggr)=7(\text{something})$$
and you can't use a variable that only exists within the limit, outside of the limit. (This is the same idea as scoping in computer programming, if you know anything about that.)
If you try to just go ahead and multiply by $x - 1$ anyway, you're using the same variable name $x$ for two different variables: one inside the limit, and one outside the limit. Here I'll use color to distinguish them:
$$(\color{red}x - 1)\biggl(\lim_{x\to 1}\frac{x^2+(3-a)x+3a}{x-1}\biggr) = 7(\color{red}x - 1)\tag{1}$$
You can bring $\color{red}x - 1$ inside the limit, because $a \lim f(x) = \lim af(x)$ (given that $a$ does not depend on $x$),
$$\lim_{x\to 1}\frac{\bigl(x^2+(3-a)x+3a\bigr)(\color{red}x - 1)}{x-1} = 7\color{red}x - 7$$
but you can't cancel out $\color{red}x - 1$ with $x - 1$ because they're different variables. In general, you should just give them different names: instead of $\color{red}x$, use $y$, for example. Then you get
$$\lim_{x\to 1}\frac{\bigl(x^2+(3-a)x+3a\bigr)(y - 1)}{x-1} = 7y - 7$$
and in that case it's pretty clear why you can't cancel anything.
However, something you can do (which is a little sneaky if you think about it, but not complicated) is use the multiplicative property of limits, namely that
$$\lim_{y\to a}f(y)\lim_{z\to a}g(z) = \lim_{x\to a}f(x)g(x)\tag{2}$$
It's usually written using the same letter for the variable in each factor, but technically they are different variables so I've made that explicit. Looking back at equation (1), instead of bringing $\color{red}x - 1$ inside the limit, you can take another limit
$$\begin{align}
\biggl(\lim_{\color{red}x\to 1}\color{red}x - 1\biggr)\biggl(\lim_{x\to 1}\frac{x^2+(3-a)x+3a}{x-1}\biggr) &= \lim_{\color{red}x\to 1}7(\color{red}x - 1) \\
\lim_{\color{blue}x\to 1}(\color{blue}x - 1)\frac{\color{blue}x^2+(3-a)\color{blue}x+3a}{\color{blue}x-1} &= 0
\end{align}$$
where on the left side I've used the multiplicative property (2), and this time you can cancel out the factors of $\color{blue}x - 1$ to get
$$\lim_{\color{blue}x\to 1}\color{blue}x^2+(3-a)\color{blue}x+3a = 1+(3-a)+3a = 0$$
This is the same result as in Ivo Terek's answer, just achieved using different (more verbose) reasoning. In this explicit reasoning the step where you take the second limit in $\color{red}x$ is key.
A: You can multiply by, say, $y-1$ but you can't push it inside the limit. The value $x$ with respect to the limit lives within the context of the limit, not external to it.
A: Consider this(not exactly and only for one side):
$$
\lim_{x\to 1} \dfrac{x^2+(3-a)x+3a}{x-1}=7\\
\dfrac{(1.00001)^2+(3-a)(1.00001)+3a}{(1.00001)-1}=7\\
(1.00001)^2+(3-a)(1.00001)+3a=7((1.00001)-1)$$
Does that make sense?
A: $$\lim_{x\to 1}\dfrac{x^2+(3-a)x+3a}{x-1}=7$$
$$\lim_{x\to 1}\dfrac{x^2+(3-a)x+3a}{x-1}=\lim_{x\to 1}7$$
$$\lim_{x\to 1}\dfrac{x^2+(3-a)x+3a}{x-1}-\lim_{x\to 1}7=0$$
$$\lim_{x\to 1}\left(\dfrac{x^2+(3-a)x+3a}{x-1}-7\right)=0$$
$$\lim_{x\to 1}\dfrac{x^2+(3-a)x+3a-7x+7}{x-1}=0$$
$$\lim_{x\to 1}\dfrac{x^2+(-4-a)x+3a+7}{x-1}=0$$
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 $a=-2$
A: No, you can not multiply by $x - 1$, because it is part of the limit operation.  This is similar to saying that in the equation
$$\sin \frac{1}{x - 1} = 1$$
you cannot solve it by multiplying by $x - 1$:
$$\sin 1 = x - 1 \implies x = 1 + \sin 1 \quad \text{(not!)}$$
However, in your situation it is additionally true that $x$ is a "dummy variable".  Your equation is for $a$; $x$ does not really exist.
A: Since none of the 10 answers so far really states this clearly, I'll expand my comment to an answer.
Your multiplication is valid for the implications $\implies$ that you wrote (but in the second line you should have written "$\lim_{x\to1}$" on the right as well as on the left). Indeed in general if $\lim_{x\to a}f(x)=c$, and if $\lim_{x\to a}g(x)$ exists (a finite value), then it must be the case that also $\lim_{x\to a}f(x)g(x)=\lim_{x\to a}cg(x)$; you applied this for the special case $a=1$ and $g(x)=x-1$. This however only shows that the original equation cannot be valid unless $a=-2$.
However, you have not shown that the original equation is actually valid for $a=-2$ (the opposite implication). In fact you would have obtained the conclusion $a=-2$ in this way even if you replace $7$ by any other number (try it!), so you certainly have not shown that the limit is $7$ for $a=-2$.
Back to the general situation of my first paragraph, if you want to deduce in the opposite direction from $\lim_{x\to a}f(x)g(x)=\lim_{x\to a}cg(x)$ that $\lim_{x\to a}f(x)=c$, then this amounts to multiplying both sides by $\frac1{g(x)}$, and this is only allowed if $\lim_{x\to a}\frac1{g(x)}$ exists. That is not the case for $a=1$ and $g(x)=x-1$.
Just for clarity, here is a derivation that regardless of the RHS, one must have $a=-2$ for the initial statement to be true. However, this condition is only necessary, not always sufficient.
$$
  \begin{align*}
&\lim_{x\to 1} \dfrac{x^2+(3-a)x+3a}{x-1}=c\\
\implies&\lim_{x\to 1} x^2+(3-a)x+3a=\lim_{x\to 1}c(x-1)\\
\iff&1^2+(3-a)\times1+3a=0c \quad\text{(both sides are continuous at $x=1$)}\\
\iff&4+2a=0\\
\iff& a=-2
\end{align*}
$$
In fact one computes for $a=-2$ that 
$$
  \lim_{x\to 1} \frac{x^2+(3-a)x+3a}{x-1}=\lim_{x\to 1}x^2+5x+6=\lim_{x\to 1}(x+6)=7,$$
so the condition is sufficient only for $c=7$.
A: yes you can multiply it but you don't let to enter that into limit and simply limit.
A: You could also solve this in the following way: since the denominator goes to zero for $x\rightarrow1$, in order for this limit to have a finite value, also the numerator should go to zero (otherwise you would get some value divided by something that goes to zero, so that the limit is $\pm\infty$).
Now you can apply l'Hôpital's rule. Taking derivatives gives:
$$\frac{2x+(3-a)}{1}=7 \quad\Longrightarrow\quad a=-2.$$
Of course you should now check that this is indeed a root of the original numerator. If it's not, then there's no solution. In this case it's alright.
A: '$\displaystyle\lim_{x\to 1}$' is a function on expressions with the variable $x$ and generally it is not true that $a\cdot f(t)=f(a\cdot t)$. It has to be proved for each function $f$.
A: You certainly can multiply a limit by a polynomial. Formally, polynomials are continuous, so $$lim_{x \rightarrow k}f(x) = A$$, then (a fixed-up version of) $f$ is continuous in a neighbourhood of $k$, which can be multiplied by a continuous function, so
$$lim_{x \rightarrow k}f(x)P(x) = A P(k)$$
for every polynomial $P$.
However, you can't always divide by a polynomial - so you may be introducing spurious solutions - for example, your attempt would also work for $A=6$:
$$lim_{x \rightarrow 1} \frac{x^2 + (3-a)x +3a}{x-1} = 6$$
$$lim_{x \rightarrow 1} x^2 + (3-a)x +3a = 6(x-1)|_{x=1}$$
$$1^2 + (3-a)1 + 3a = 0$$
$$a = -2$$
but certainly $$lim_{x \rightarrow 1} \frac{x^2 + (3-(-2))x +3(-2)}{x-1} \ne 6$$
This is similar to what happens when you multiply out an rational equation ("algebraic fraction") or square both sides of an equation - you just have to check the solutions you found.
