The limit of $x$ square minus four divided by $x$ minus $2$ is undefined as the value of $x$ approaches two when we try to evaluate the limit by the direct substitution method.
$\displaystyle \large \lim_{x\,\to\,2}{\normalsize \dfrac{x^2-4}{x-2}}$ $\,=\,$ $\dfrac{0}{0}$
The expression in the numerator $x$ squared minus four is a second degree polynomial and it can be factored. So, let us try to find the limit of square of $x$ minus $4$ divided by $x$ minus $2$ by the factoring method as the value $x$ closer to $2$.
The first term in the numerator is in square form. The input of limit operation tends to $2$ and the second term in the denominator of rational function is also $2$. So, it is recommended to express the second term in the numerator in terms of $2$ and also in square form.
$=\,\,$ $\displaystyle \large \lim_{x\,\to\,2}{\normalsize \dfrac{x^2-2 \times 2}{x-2}}$
The number $2$ is multiplied by itself. So, let’s express the product of the factors in exponential notation.
$=\,\,$ $\displaystyle \large \lim_{x\,\to\,2}{\normalsize \dfrac{x^2-2^2}{x-2}}$
Now, the difference of two squares can be factored by the difference of squares rule.
$=\,\,$ $\displaystyle \large \lim_{x\,\to\,2}{\normalsize \dfrac{(x+2)(x-2)}{x-2}}$
The algebraic expression in the numerator is simplified as a product of two factors and there is nothing to simplify in the denominator. So, it is time to concentrate on simplifying the rational function.
$=\,\,$ $\displaystyle \large \lim_{x\,\to\,2}{\normalsize \dfrac{(x+2) \times (x-2)}{x-2}}$
$=\,\,$ $\displaystyle \large \lim_{x\,\to\,2}{\normalsize \dfrac{(x+2) \times \cancel{(x-2)}}{\cancel{x-2}}}$
$=\,\,$ $\displaystyle \large \lim_{x\,\to\,2}{\normalsize (x+2)}$
The rational function $x$ square minus $4$ divided by $x$ minus $2$ is simplified as an algebraic expression $x$ plus $2$, and the simplified algebraic expression cannot be simplified further. So, the limit of the function can be now evaluated by the direct substitution method.
$=\,\,$ $2+2$
$=\,\,$ $4$
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