If $\lim_{x\to1}\frac{f(x)-8}{x-1}=10$, then find $\lim_{x\to1}f(x)$.
My solution was as follows:
\begin{align}
\lim_{x\to1}\frac{f(x)-8}{x-1}=10&\Leftrightarrow \frac{\lim_{x\to1}f(x)-\lim_{x\to1}8}{\lim_{x\to1}x-\lim_{x\to1}1}=10\\[1em]
&\Leftrightarrow \lim_{x\to1}f(x)-\lim_{x\to1}8=10(\lim_{x\to1}x-\lim_{x\to1}1)\\[1em]
&\Leftrightarrow \lim_{x\to1}f(x)=10(\lim_{x\to1}x-\lim_{x\to1}1)+\lim_{x\to1}8\\[1em]
&\Leftrightarrow \lim_{x\to1}f(x)=10(1-1)+8=8.
\end{align}
Hence, $\lim_{x\to1}f(x)=8$.
The book, however, gives the following:
$$
\lim_{x\to1}[f(x)-8]=\lim_{x\to1}\left[\frac{f(x)-8}{x-1}\cdot(x-1)\right]=\lim_{x\to1}\frac{f(x)-8}{x-1}\cdot\lim_{x\to1}(x-1)=10\cdot0=0.
$$
Thus,
$$
\lim_{x\to1}f(x)=\lim_{x\to1}\{[f(x)-8]+8\}=\lim_{x\to1}[f(x)-8]+\lim_{x\to1}8=0+8=8.
$$
There is then a remark about how the value of $\lim_{x\to1}\frac{f(x)-8}{x-1}$ does not affect the answer since it's multiplied by 0. What is important is that the limit exists.
Question: Is there anything wrong with my answer? It seems a lot easier / more efficient.
Best Answer
Here's another solution since I'm not too big of a fan on symbolic limits.
Since \begin{align} \lim_{x\rightarrow 1}\frac{f(x)-8}{x-1} = 10 \end{align} then it follows \begin{align} \lim_{x\rightarrow 1} \frac{f(x)-8-10(x-1)}{x-1} = 0. \end{align} which means when $x$ is close to $1$ we have have that \begin{align} \left|\frac{f(x)-8 -10(x-1)}{x-1} \right| \leq 1\ \ \Rightarrow\ \ |f(x)-8-10(x-1)| \leq |x-1|. \end{align} Using triangle inequality, we have that \begin{align} |f(x)-8| \leq 11|x-1|. \end{align} Thus, it follows $\lim_{x\rightarrow 1} f(x) = 8$.