How to solve this limit using only limit laws? [$\lim_{x\to 2} {f(x) -5 \over x-2} = 3 $]

Question

$$ \lim_{x\to 2} {f(x) -5 \over x-2} = 3 . \quad \text{Find} \lim_{x \to 2}f(x). $$

In the following limit, my initial thought process was to use the limit law: $$ \lim_{x\to c} {g(x) \over h(x)} = {\lim_{x \to c} g(x) \over \lim_{x \to c} h(x)} $$

But this only applies when $\lim_{x \to c} h(x) \not= 0$ if I'm correct. In this case however, that doesn't hold so I shouldn't be able to use this to find $f(x)$. But then how else am I supposed to solve this.

My working was this: $$ \lim_{x\to 2} {f(x) -5 \over x-2} = 3 \\ \implies \lim_{x \to 2} f(x) -5 = 0 \\ \implies \lim_{x \to 2} f(x) = 5 $$

This is an exercise question from Thomas' Calculus (12 ed.) from the limits and continuity chapter. And the answer for this exercise is apparently correct, but this feels very wrong and something doesn't fit in.

Answer 1

The limit can only be finite if $\displaystyle \lim_{x\to 2}(f(x)-5) = 0$. You can characterize $f$ as $$ f(x) - 5 = \varphi(x) (x-2) \Leftrightarrow f(x)=5+\varphi(x)(x-2), $$

with $\varphi(2) = 3$.

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As it turns out, the original exercise only asked for $\displaystyle \lim_{x\to 2}f(x)$...

Answer 2

$$ \lim_{x\to 2} {f(x) -5 \over x-2} = 3 \quad $$ Set $f(2)-5=0$ to apply the L Hospital rule: $$ \lim_{x\to 2} {f'(x)} = 3 $$ We must have $f'(2)=3$, so $f'(x)$ should be differentiable at $x=2$

Thus, any general function with the property $f(2)=5$ and $f'(2)=3$ will do.

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As mentioned by zwim, the definition of the derivative is formally used to realize the conditions above.

By setting $f(2)=5$:

$$ f'(2)=\lim_{x\to 2}{f(x)-f(2)\over x-2}=3$$

Again, we see the sufficient conditions:

$$f(2)=5$$ $$f'(2)=3$$

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For example, the functions: $$f(x)=3x-1$$ $$f(x)=x^2-x+3$$ $$ f(x) ={e^{3x}\over e^6}+4$$

satisfy both the conditions above