calculating a function in a limit expression

Question

In the exercises of Section 1.2 in Adams Calculus, I came across the following question (no l'Hopital allowed!).

If $\displaystyle\lim_{x\mapsto2}\frac{f(x)-5}{x-2}=3$ then find $\displaystyle\lim_{x\mapsto2}f(x)$.

I am stumped. The "obvious" thing is to start by going $\cdots=\frac{\lim_{x\mapsto2}(f(x)-5)}{\lim_{x\mapsto2}(x-2)}=3$, but this is clearly wrong. Multiplying the whole fraction by a rational function $p(x)/p(x)$ in an attempted to get rid of the $x-2$ by cancelling with something from the top line is doomed to failure, as any cancelling might as well happen in the $p(x)$ rather than the $x-2$.

Help?

Answer 1

Let

$$g(x)=\frac{f(x)-5}{x-2}-3$$ then we have by hypothesis $$\lim_{x\to2}g(x)=0$$ Moreover, we have

$$f(x)=5+(x-2)(g(x)+3)$$ Can you now answer the question?

Answer 2

Hint

$f(x)-5$ has be in the form $3(x-2)g(x)$, with $\lim_{x \to 2} g(x) = 1$ or the limit wouldn't exist.

Answer 3

Hint: Suppose that the limit is $\lambda$ and goes forward from there, using what you know. What can you deduce on $\lambda$?