Question

Solve \[\underset{x\to 0}{\mathop \lim }\,\dfrac{{{\left( 27+x \right)}^{\dfrac{1}{3}}}-3}{9-{{\left( 27+x \right)}^{\dfrac{2}{3}}}}\]?
A. \[-\dfrac{1}{3}\]
B. \[\dfrac{1}{6}\]
C. \[-\dfrac{1}{6}\]
D. \[\dfrac{1}{3}\]

Answer 1

Hint: In the given question, we have been asked to find the value of the given expression i.e. \[\dfrac{{{\left( 27+x \right)}^{\dfrac{1}{3}}}-3}{9-{{\left( 27+x \right)}^{\dfrac{2}{3}}}}\] and it is also given that the value of x approaches to 0. In order to solve the given expression, first we will take out 3 as a common term from numerator and 9 as common term from denominator. Later by simplifying the numbers using the mathematical operations such as addition, subtraction, multiplication and division and also by cancelling out the common terms from the numerator and denominator. We need to simplify the given expression till we don't get any value.

Complete step by step solution:
We have given that,
\[ \underset{x\to 0}{\mathop \lim }\,\dfrac{{{\left( 27+x \right)}^{\dfrac{1}{3}}}-3}{9-{{\left( 27+x \right)}^{\dfrac{2}{3}}}}\]
Taking out 3 as a common terms from the numerator and 9 from the denominator,
We will obtain,
\[\Rightarrow \underset{x\to 0}{\mathop \lim }\,\dfrac{3\left( {{\left( 1+\dfrac{x}{27} \right)}^{\dfrac{1}{3}}}-1 \right)}{9\left( 1-{{\left( 1+\dfrac{x}{27} \right)}^{\dfrac{2}{3}}} \right)}\]
Cancelling out common factors, we will get
\[\Rightarrow \underset{x\to 0}{\mathop \lim }\,\dfrac{1\left( {{\left( 1+\dfrac{x}{27} \right)}^{\dfrac{1}{3}}}-1 \right)}{3\left( 1-{{\left( 1+\dfrac{x}{27} \right)}^{\dfrac{2}{3}}} \right)}\]
Simplifying the above expression, we will get
\[\Rightarrow \underset{x\to 0}{\mathop \lim }\,\dfrac{1\left( \left( 1+\dfrac{1}{3}\times \dfrac{x}{27} \right)-1 \right)}{3\left( 1-\left( 1+\dfrac{2}{3}\times \dfrac{x}{27} \right) \right)}\]
Simplifying the above expression, we will get
\[\Rightarrow \underset{x\to 0}{\mathop \lim }\,\dfrac{1\left( \left( 1+\dfrac{x}{81} \right)-1 \right)}{3\left( 1-\left( 1+\dfrac{2x}{81} \right) \right)}\]
Solving the brackets in the numerator and the denominator, we will get
\[\Rightarrow \underset{x\to 0}{\mathop \lim }\,\dfrac{1\left( \left( 1+\dfrac{x}{81} \right)-1 \right)}{3\left( 1-\left( 1+\dfrac{2x}{81} \right) \right)}=\underset{x\to 0}{\mathop \lim }\,\dfrac{1\times \dfrac{x}{81}}{3\times -\dfrac{2x}{81}}\]
Simplifying the numbers in the numerators and denominators, we will get
\[\Rightarrow \underset{x\to 0}{\mathop \lim }\,\dfrac{1\times \dfrac{x}{81}}{3\times -\dfrac{2x}{81}}=\underset{x\to 0}{\mathop \lim }\,\dfrac{\dfrac{x}{81}}{-\dfrac{6x}{81}}=\underset{x\to 0}{\mathop \lim }\,\dfrac{x}{81}\times -\dfrac{81}{6x}\]
Cancelling out the common terms, we will get
\[\Rightarrow \underset{x\to 0}{\mathop \lim }\,\dfrac{x}{81}\times -\dfrac{81}{6x}=-\dfrac{1}{6}\]
Therefore,
\[\Rightarrow \underset{x\to 0}{\mathop \lim }\,\dfrac{{{\left( 27+x \right)}^{\dfrac{1}{3}}}-3}{9-{{\left( 27+x \right)}^{\dfrac{2}{3}}}}=-\dfrac{1}{6}\]

Hence, the option (C) is the correct answer.

Note: While solving these types of questions, students need to do the calculation part very carefully to avoid making any error. To solve these types of questions all you need to know is to perform the mathematical operations such as addition, subtraction, multiplication and division.
We can also solve this question using L’s hospital rule as the given expression is \[\dfrac{0}{0}\]form, when we put the limit in the given function.