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I'm a first year student of Mathematics. Obviously, this is not a undergraduate Mathematics question and I also know its answer which is $\frac{1}{4}$. But I have a doubt regarding the solution of that problem. I have see many people(even I also when I am in my higher school) to solve the problem as follows:
$\lim_{x\to 2}\frac{x-2}{x^2-4}$
$=\lim_{x\to 2}\frac{x-2}{(x+2)(x-2)}$
$=\lim_{x\to 2}\frac{1}{x+2}$ $($$(x-2)$ is cancelled out from numerator and denominator since $x\neq2$$)$
$=\frac{1}{2+2}$
$=\frac{1}{4}$
But in the second line(third from last) for the part $\frac{x-2}{x-2}$ they write $x\neq2$ but in the last line for the part $\frac{1}{x+2}$ they ultimately put $x=2$ ALTHOGH BOTH THE TERMS $\frac{x-2}{x-2}$ AND $\frac{1}{x+2}$ ARE INSIDE THE LIMIT.
So, I think there is a little hidden fallacy inside this proof although I have seen this solution in many places especially in high school.
I am unable to find out this hidden problem more analytically.
Can anybody give me a rigorous and more analytical way out to find where the actual problem lies and also a rigorous solution to the limit?
N.B. I have avoided the solution by L'Hôpital Rule. And also Thanks for your soluton in advance.

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  • $\begingroup$ I think it make sense if you draw both graphs to see the difference between $\frac {x-2}{x^2-4}$ and $\frac{1}{x+2}$, at $x=2$. $\endgroup$ Commented Apr 2, 2018 at 19:29
  • $\begingroup$ @BiswarupSaha Please remember that you can choose an answer among the given if the OP is solved, more details here meta.stackexchange.com/questions/5234/… $\endgroup$
    – user
    Commented Apr 7, 2018 at 19:45

4 Answers 4

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Your question is on the nature of limits, rather than on the calculation itself. The existence and value of $\lim_{x\to a} f(x)$ do not depend on the value of $f(a)$ -- in fact, they do not depend on the existence of $f(a)$. They depend only on the value of $f$ near $a$.

When $f$ is continuous, it happens that $\lim_{x\to a}f(x)$ equals $f(a)$ if $a$ lies on the domain $f$.


In particular, if $f$ and $g$ agree near $a$, even if one or both of them are not defined at $x=a$, then the limits $\lim_{x\to a}f(x)$ and $\lim_{x\to a}g(x)$ either both exist and are equal, or both do not exist (and if one is infinite, so is the other, with the same sign).

If in addition to $f$ and $g$ agreeing near $a$ we also have that $g$ is continuous and $a$ lies on the domain of $g$, then the limits exist and

$$\lim_{x\to a}f(x) = g(a),$$

once again, even if $a$ does not lie on the domain of $f$.

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  • $\begingroup$ This is only half of the answer because the initial function is undefined at $x=2$. $\endgroup$
    – user65203
    Commented Apr 2, 2018 at 19:51
  • $\begingroup$ How exactly is this half of the answer? $\endgroup$ Commented Apr 2, 2018 at 19:59
  • $\begingroup$ There are two different functions in this problem. $\endgroup$
    – user65203
    Commented Apr 2, 2018 at 20:03
  • $\begingroup$ Yes, which coincide near $a=2$. $\endgroup$ Commented Apr 2, 2018 at 20:04
  • $\begingroup$ This is what is missing in your answer. $\endgroup$
    – user65203
    Commented Apr 2, 2018 at 20:04
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The correct way is

$$\lim_{x\to 2}\frac{x-2}{x^2-4}=\lim_{x\to 2}\frac{x-2}{(x-2)(x+2)}=\lim_{x\to 2}\frac{1}{x+2}=\frac14$$

Regarding this kind of cancelation refer to Why are we allowed to cancel fractions in limits?.

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You can substitute $x=2$ because the function inside the limit is continuous at $x=2$, so the limit equals to its value at that point.

Another way: $$\lim_{x\to2} \frac{x-2}{x^2-4}=\lim_{x\to2}\frac{x-2}{(x-2)(x+2)}=\lim_{x\to2}\frac{x-2}{x-2}\lim_{x\to2}\frac{1}{x+2}=1\lim_{x\to2}\frac{1}{x+2}$$

But of course, you can also use the $\epsilon$-$\delta$ definition of the limit.

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  • $\begingroup$ Yes, but what about the ambiguity in the second line(third line from last) where we put $x$ to be not equal to $2$ $\endgroup$
    – MathBS
    Commented Apr 2, 2018 at 19:28
  • $\begingroup$ @BiswarupSaha Because $x \neq 2$. We are just studying the behaviour of the function around $2$. So we can simplify by $x-2$. And then, you are left with another function, which is continuous at $x=2$, so it's limit at $x=2$ equals to its value. $\endgroup$
    – Botond
    Commented Apr 2, 2018 at 19:32
  • $\begingroup$ @BiswarupSaha I added a different approach to the answer, using the product rule for limits. $\endgroup$
    – Botond
    Commented Apr 2, 2018 at 19:34
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For all $x\ne2$,

$$\frac{x-2}{x^2-4}=\frac1{x+2}$$

and this allows you to write

$$\lim_{x\to2}\frac{x-2}{x^2-4}=\lim_{x\to2}\frac1{x+2}.$$

Then the evaluation of the second limit

$$\lim_{x\to2}\frac1{x+2}=\frac1{2+2}$$

is correct because the function is continuous, so that $\lim_{x\to 2}f(x)=f(2)$. You can establish this by an $\epsilon-\delta$ argument:

$$\left|\frac1{2+\epsilon+2}-\frac14\right|=\frac{\epsilon}{4+\epsilon}<\epsilon.$$

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