Question: The question is:

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> Suppose $\lim\_{x\rightarrow 1}\frac{f(x)-7}{x-1}=4$, find $\lim\_{x\rightarrow 1}f(x)$.
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The obvious answer is $7$, by going:
$$\begin{align\*}
\lim\_{x\rightarrow 1}\frac{f(x)-7}{x-1}&=4\\
\Rightarrow\lim\_{x\rightarrow 1}[f(x)-7]&=4\times{\lim\_{x\rightarrow 1}[x-1]}\\
\Rightarrow\lim\_{x\rightarrow 1}[f(x)-7]&=0\\
\Rightarrow\lim\_{x\rightarrow 1}f(x)&=7
\end{align\*}$$
But this feels wrong. Specifically, the step of taking "${\lim\_{x\rightarrow 1}[x-1]}$" over to the right hand side seems illegal. I justify this to myself by saying we can treat it like a number not equal to zero because it is a limit. But I don't believe my justification.

Answer: 
**Hint**: Its a matter of rewrite $f(x) = (x-1)\cdot \dfrac{f(x)-7}{x-1} + 7\Rightarrow \displaystyle \lim\_{x\to 1} f(x) = \displaystyle \lim\_{x\to1}(x-1)\cdot \displaystyle \lim\_{x\to 1}\dfrac{f(x)-7}{x-1} + \displaystyle \lim\_{x \to 1} 7 = ?$
The subtlety of this question is where the 4 comes from, since $f(1)$ must be 7 for the limit to exist. This quotient is the definition of *derivitive* so what it's saying is that $f'(1) = 4.$