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Find $\displaystyle \large \lim_{x\,\to\,2}{\normalsize \dfrac{x^3-8}{x^2-4}}$ by formulas

According to the direct substitution method, it has evaluated that the limit of the quotient of x cube minus 8 by x square minus 4 is indeterminate as the value of $x$ tends to $2$. Hence, the limit of this rational function is calculated by factorization or factorisation.

$\displaystyle \large \lim_{x\,\to\,2}{\normalsize \dfrac{x^3-8}{x^2-4}}$

The limit of cube of $x$ minus $8$ by square of $x$ minus $4$ can be evaluated as the value of $x$ approaches $2$ by using a standard result of limits.

Adjust the rational function to standard result

The constants in the expressions are $8$ and $4$, and they can be expressed in exponential notation in terms of the input of limit operation.

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,2}{\normalsize \dfrac{x^3-2^3}{x^2-2^2}}$

The limit of the rational function matches with power difference limit rule in ratio form. So, it is recommendable to adjust this rational function exactly same as the limit rule.

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,2}{\normalsize \bigg(\dfrac{x^3-2^3}{x^2-2^2} \times 1\bigg)}$

It is essential to include an expression $x-2$ in the rational function in order to convert the given rational function into the required form as per the power difference limit rule in ratio form.

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,2}{\normalsize \bigg(\dfrac{x^3-2^3}{x^2-2^2} \times \dfrac{x-2}{x-2}\bigg)}$

Now, let us try to express the product of rational functions into power difference limit rule in ratio form as per the multiplication of the fractions.

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,2}{\normalsize \dfrac{(x^3-2^3) \times (x-2)}{(x^2-2^2) \times (x-2)}}$

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,2}{\normalsize \dfrac{(x^3-2^3) \times (x-2)}{(x-2) \times (x^2-2^2)}}$

Now, split the rational function as a product of two rational functions.

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,2}{\normalsize \bigg(\dfrac{x^3-2^3}{x-2} \times \dfrac{x-2}{x^2-2^2}\bigg)}$

Use the product rule of limits to find the limit of product of two functions by the product of their limits.

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,2}{\normalsize \dfrac{x^3-2^3}{x-2}}$ $\times$ $\displaystyle \large \lim_{x\,\to\,2}{\normalsize \dfrac{x-2}{x^2-2^2}}$

Evaluate the Limit of each rational function

The first factor represents the power difference limit rule in ratio form but the rational function in the second factor is in reciprocal form. However, it can be written in its multiplicative inverse form as follows.

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,2}{\normalsize \dfrac{x^3-2^3}{x-2}}$ $\times$ $\displaystyle \large \lim_{x\,\to\,2}{\normalsize \dfrac{1}{\dfrac{x^2-2^2}{x-2}}}$

According to the reciprocal rule of limits, the limit of the reciprocal of a function can be calculated by the reciprocal of its limit.

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,2}{\normalsize \dfrac{x^3-2^3}{x-2}}$ $\times$ $\dfrac{1}{\displaystyle \large \lim_{x\,\to\,2}{\normalsize \dfrac{x^2-2^2}{x-2}}}$

Now, find the limit of each rational function as per the power different limit rule in ratio form formula.

$=\,\,\,$ $3 \times 2^{3-1}$ $\times$ $\dfrac{1}{2 \times 2^{2-1}}$

Simplify the expression to find the limit

Now, simplify this arithmetic expression to find the limit of the given rational function as the value of $x$ is closer to $2$.

$=\,\,\,$ $3 \times 2^{2}$ $\times$ $\dfrac{1}{2 \times 2^{1}}$

$=\,\,\,$ $3 \times 4$ $\times$ $\dfrac{1}{2 \times 2}$

$=\,\,\,$ $12$ $\times$ $\dfrac{1}{4}$

$=\,\,\,$ $\dfrac{12 \times 1}{4}$

$=\,\,\,$ $\dfrac{12}{4}$

$=\,\,\,$ $\dfrac{\cancel{12}}{\cancel{4}}$

$=\,\,\,$ $3$

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