# Limit question

#### Amer

##### Active member
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

#### Opalg

##### MHB Oldtimer
Staff member
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

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.