Question

Solve: $\mathop {\lim }\limits_{x \to 27} \dfrac{{({x^{\dfrac{1}{3}}} + 3)({x^{\dfrac{1}{3}}} - 3)}}{{x - 27}}$

Answer 1

In the above question, if we directly put the limit in place of the variable, then the denominator will result in zero and hence the solution will result is not defined. So first we have to simplify the given expression by applying a suitable formula and then we will cancel out the like terms and after that, we will put the limits.

Complete step by step answer:
If we put the given limit in the function in place of the variable and if both the numerator and the denominator results in zero, then we will first simplify the expression and then we will further solve our question.
In the given question, we have to solve $\mathop {\lim }\limits_{x \to 27} \dfrac{{({x^{\dfrac{1}{3}}} + 3)({x^{\dfrac{1}{3}}} - 3)}}{{x - 27}}$
If we put a limit here then the denominator comes out zero so we will get an undefined value. So first we will simplify the above expression.
$\mathop {\lim }\limits_{x \to 27} \dfrac{{({x^{\dfrac{1}{3}}} + 3)({x^{\dfrac{1}{3}}} - 3)}}{{{{({x^{\dfrac{1}{3}}})}^3} - {{(3)}^3}}}$ …….. ($1$)
In the denominator of the above expression, x can be written as ${({x^{\dfrac{1}{3}}})^3}$ and $27$ can be written as ${(3)^3}$.
We know that, according to the formula given below.
${a^3} - {b^3} = (a - b)({a^2} + ab + {b^2})$
Now we will apply this formula in the denominator of the first equation.
${({x^{\dfrac{1}{3}}})^3} - {(3)^3} = ({x^{\dfrac{1}{3}}} - 3)({x^{\dfrac{2}{3}}} + 9 + 3{x^{\dfrac{1}{3}}})$ ……….. ($2$)
On putting the value of ($2$) in ($1$), we get the following results.
$\mathop {\lim }\limits_{x \to 27} \dfrac{{({x^{\dfrac{1}{3}}} + 3)({x^{\dfrac{1}{3}}} - 3)}}{{({x^{\dfrac{1}{3}}} - 3)({x^{\dfrac{2}{3}}} + 9 + 3{x^{\dfrac{1}{3}}})}}$
The value $({x^{\dfrac{1}{3}}} - 3)$ will cancel out each other in the numerator and the denominator and we get the results as shown below.
$\mathop {\lim }\limits_{x \to 27} \dfrac{{({x^{\dfrac{1}{3}}} + 3)}}{{({x^{\dfrac{2}{3}}} + 9 + 3{x^{\dfrac{1}{3}}})}}$ ….. ($3$)
We know that $\mathop {\lim }\limits_{x \to 27}$ means that we have to place the value of $27$ in place of x. On applying limits in (3), we get the results as shown below.
$\mathop {\lim }\limits_{x \to 27} \dfrac{{({x^{\dfrac{1}{3}}} + 3)}}{{({x^{\dfrac{2}{3}}} + 9 + 3{x^{\dfrac{1}{3}}})}}$,
$\Rightarrow \dfrac{{({{(27)}^{\dfrac{1}{3}}} + 3)}}{{({{(27)}^{\dfrac{2}{3}}} + 9 + 3{{(27)}^{\dfrac{1}{3}}})}}$,
$27$ is the cube of $3$, so $27$ can be written as ${(3)^3}$ and the above expression becomes
$\Rightarrow \dfrac{{({{({3^3})}^{\dfrac{1}{3}}} + 3)}}{{({{({3^3})}^{\dfrac{2}{3}}} + 9 + 3{{({3^3})}^{\dfrac{1}{3}}})}}$,
$\Rightarrow \dfrac{{(3 + 3)}}{{(9 + 9 + 9)}}$,
$\Rightarrow \dfrac{6}{{27}}$,
$\Rightarrow \dfrac{2}{9}$
Hence, after solving the limit we get the following results.
$\mathop {\lim }\limits_{x \to 27} \dfrac{{({x^{\dfrac{1}{3}}} + 3)({x^{\dfrac{1}{3}}} - 3)}}{{x - 27}} = \dfrac{2}{9}$

Note:
If we put the limit in place of the variable in the function and if both the numerator and denominator result in zero then L'hospital's Rule will be applied on the given expression, according to which, first we have to differentiate both numerator and denominator expression with respect to x and then we will apply the limits.