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Limit question

Amer

Active member
Mar 1, 2012
275
I gave my students a question that said
if
[tex]\lim _{x \rightarrow 2 } \frac{f(x)}{x^2-4} = 7 [/tex]

find the limit

[tex]\lim_{x\rightarrow 2 } \frac{f(x)}{x-2 } [/tex]

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

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

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

What is your opinion ?
Thanks in advance
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,708
I gave my students a question that said
if
[tex]\lim _{x \rightarrow 2 } \frac{f(x)}{x^2-4} = 7 [/tex]

find the limit

[tex]\lim_{x\rightarrow 2 } \frac{f(x)}{x-2 } [/tex]

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

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

which is true, what he did in this
[tex]\frac{f(x)}{x^2-4} = 7 \Rightarrow f(x) = 7(x^2-4) [/tex]
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

Active member
Mar 1, 2012
275
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