MHB Opinion Needed: Is Student's Solution Around x=2 Correct?

[Amer]

I gave my students a question that said
if
\lim _{x \rightarrow 2 } \frac{f(x)}{x^2-4} = 7

find the limit

\lim_{x\rightarrow 2 } \frac{f(x)}{x-2 }

one of my students answered like this
from the given
\frac{f(x)}{x^2-4} = 7 \Rightarrow f(x) = 7(x^2-4)

then he complete the solution
\lim_{x\rightarrow 2} \frac{7(x-2)(x+2)}{x-2} = 28

which is true, what he did in this
\frac{f(x)}{x^2-4} = 7 \Rightarrow f(x) = 7(x^2-4)
is not true 100%, it is true around 2

What is your opinion ?
Thanks in advance

 

[Opalg]

I gave my students a question that said
if
\lim _{x \rightarrow 2 } \frac{f(x)}{x^2-4} = 7

find the limit

\lim_{x\rightarrow 2 } \frac{f(x)}{x-2 }

one of my students answered like this
from the given
\frac{f(x)}{x^2-4} = 7 \Rightarrow f(x) = 7(x^2-4)

then he complete the solution
\lim_{x\rightarrow 2} \frac{7(x-2)(x+2)}{x-2} = 28

which is true, what he did in this
\frac{f(x)}{x^2-4} = 7 \Rightarrow f(x) = 7(x^2-4)
is not true 100%, it is true around 2

What is your opinion ?
Thanks in advance

Of course the student was wrong to say that $f(x) = 7(x^2-4)$. But it is "approximately true" as $x\to2$ and he got the correct answer to the question. I would give him some partial credit for a method that was not too far from correct. If he had expressed it as $$\lim_{x\rightarrow 2 } \frac{f(x)}{x-2 } = \lim _{x \rightarrow 2 } \frac{f(x)(x+2)}{x^2-4} = \lim _{x \rightarrow 2 } \frac{f(x)}{x^2-4}\lim _{x \rightarrow 2 }(x+2) = 7\times 4 = 28$$ then he would have deserved full credit. What he missed was to make use of the important fact that the limit of a product is the product of the limits.

 

[Amer]

Thanks.
I was thinking in an example if he solve it with this way he will get an error, or how i can explane his mistake