Evaluate: $\lim_{x\to 2}\frac{x-2}{x^2-4}$ without using L'Hôpital Rule

Question

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.

Answer 1

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.

Answer 2

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?.

Answer 3

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:

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

Answer 4

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 $f$.

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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 $f$.