Solve the limit-
$\mathop {\lim }\limits_{x \to 2} \left( {\dfrac{{{x^5} - 32}}{{{x^3} - 8}}} \right)$

- Hint: Whenever a given limit is in the indeterminate form, L’Hospital rule is used. This is only applicable when the limit is in $\dfrac{0}{0}or\dfrac{\infty }{\infty }$ forms. The L’Hospital rule is given by-
$\mathop {\lim }\limits_{{\text{x}} \to {\text{a}}} \dfrac{{{\text{f}}\left( {\text{a}} \right)}}{{{\text{g}}\left( {\text{a}} \right)}} = \mathop {\lim }\limits_{{\text{x}} \to {\text{a}}} \dfrac{{{\text{f}}'\left( {\text{a}} \right)}}{{{\text{g}}'\left( {\text{a}} \right)}}\;if\;and\;only\;if\;{\text{f}}\left( {\text{a}} \right),\;{\text{g}}\left( {\text{a}} \right) \to 0,\infty $
Also, a few formula for differentiation required are-
$\dfrac{{{\text{d}}\left( {{{\text{x}}^{\text{n}}}} \right)}}{{dx}} = n{x^{{\text{n}} - 1}}...\left( 1 \right)$

Complete step-by-step solution -

The given limit is-
$\mathop {\lim }\limits_{x \to 2} \left( {\dfrac{{{x^5} - 32}}{{{x^3} - 8}}} \right)$
When we substitute x = 2 in this, we get-
$\dfrac{{{2^5} - 32}}{{{2^3} - 8}} = \dfrac{{32 - 32}}{{8 - 8}} = \dfrac{0}{0}\left( {indeterminate} \right)$
Hence, L’Hospital rule can be applied on this limit. Using property (1) we can write that-
$\mathop {\lim }\limits_{{\text{x}} \to 2} \left( {\dfrac{{{{\text{x}}^5} - 32}}{{{{\text{x}}^3} - 8}}} \right) = \mathop {\lim }\limits_{{\text{x}} \to 2} \left( {\dfrac{{5{{\text{x}}^4}}}{{3{{\text{x}}^2}}}} \right)$
This expression obtained is now a determinate form, so we can directly substitute x = 2 and get the value as-
$\mathop {\lim }\limits_{{\text{x}} \to 2} \left( {\dfrac{{5{{\text{x}}^4}}}{{3{{\text{x}}^2}}}} \right) = \dfrac{{5 \times {2^4}}}{{3 \times {2^2}}} = \dfrac{{20}}{3}$

Note: L’Hospital rule is the simplest and the shortest method to solve a limit in an indeterminate form. In this question, we can also use a factorization method which is quite lengthy, because there are larger powers of x involved. If smaller powers are x are involved, we can use the factorization method. For example-
$\mathop {\lim }\limits_{{\text{x}} \to {\text{a}}} \left( {\dfrac{{{{\text{x}}^2} - {{\text{a}}^2}}}{{{\text{x}} - {\text{a}}}}} \right) = \mathop {\lim }\limits_{{\text{x}} \to {\text{a}}} \left( {\dfrac{{\left( {{\text{x}} - {\text{a}}} \right)\left( {{\text{x}} + {\text{a}}} \right)}}{{{\text{x}} - {\text{a}}}}} \right) = \mathop {\lim }\limits_{{\text{x}} \to {\text{a}}} \left( {{\text{x}} + {\text{a}}} \right) = 2{\text{a}}$