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Let \[\alpha \left( a \right)\] and \[\beta \left( a \right)\] be the roots of the equation \[\left( {\sqrt[3]{{1 + a}} - 1} \right){x^2} + \left( {\sqrt {1 + a} - 1} \right)x + \left( {\sqrt[6]{{1 + a}} - 1} \right) = 0\] where \[a > - 1\]then, \[\mathop {\lim }\limits_{a \to {0^ + }} \alpha \left( a \right)\] and \[\mathop {\lim }\limits_{a \to {0^ + }} \beta \left( a \right)\]
A) $ - \dfrac{5}{2}and\;1$
B) $ - \dfrac{1}{2}and\; - 1$
C) $ - \dfrac{7}{2}and\;2$
D) $ - \dfrac{9}{2}and\;3$

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Last updated date: 23rd Jul 2024
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Answer
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Hint: A limit is basically a number attained by a function when the independent variable of the function approaches a given value. For example, if we have the function \[f\left( x \right){\text{ }} = {\text{ }}6x\], then it is stated as, “the limit of the function \[f\left( x \right)\] as \[\;x\] approaches 2 is 12. Mathematically, it is expressed as \[\mathop {lim}\limits_{ \to 2} f\left( x \right){\text{ }} = {\text{ }}12\].
 Continuity of a Function
A function f is said to be continuous at the point \[x{\text{ }} = {\text{ }}a\] if the following conditions are true:
\[f\left( a \right)\] is defined
\[\mathop {lim}\limits_{ \to {\text{ }}a} {\text{ }}f\left( x \right)\] exists
\[\mathop {lim}\limits_{ \to {\text{ }}a} f\left( x \right) = f(a)\]
Both side limits must also be equal i.e. \[\mathop {lim}\limits_{x \to {\text{ }}{a^ - }} {\text{ }}f\left( x \right) = f(a) = \mathop {lim}\limits_{x \to {\text{ }}{a^ + }} {\text{ }}f\left( x \right)\].

Complete step-by-step answer:
Here, we cannot put limits directly into this expression as it stands. But we can use binomial expansion, as follows;
\[\left( {\sqrt[3]{{1 + a}} - 1} \right){x^2} + \left( {\sqrt {1 + a} - 1} \right)x + \left( {\sqrt[6]{{1 + a}} - 1} \right) = 0\]
We want to use binomial theorem, we need to make sure it is in the proper form to use the Binomial Series. Here is the proper form for this function,
\[ \Rightarrow \left( {{{\left( {1 + a} \right)}^{\dfrac{1}{3}}} - 1} \right){x^2} + \left( {{{\left( {1 + a} \right)}^{\dfrac{1}{2}}} - 1} \right)x + \left( {{{\left( {1 + a} \right)}^{\dfrac{1}{6}}} - 1} \right) = 0\]
Using binomial theorem and expanding. (Substituting $n = \dfrac{1}{3}$​, , $n = \dfrac{1}{2}$ and $n = \dfrac{1}{6}$ into the binomial series, we get::-
\[ \Rightarrow \left( {1 + \dfrac{a}{3} - 1} \right){x^2} + \left( {1 + \dfrac{a}{2} - 1} \right)x + \left( {1 + \dfrac{a}{6} - 1} \right) = 0\](Now all we need to do is plug into the formula from the notes and write down the first terms. We have ignored all other terms of the binomial series as other terms were very small so we assumed them negligible.)
\[ \Rightarrow a\left( {\dfrac{{{x^2}}}{3} + \dfrac{x}{2} + \dfrac{1}{6}} \right) = 0\]
\[ \Rightarrow 2{x^2} + 3x + 1 = 0\]
Factorising the above equation.
\[ \Rightarrow \left( {2x + 1} \right)(x + 1) = 0\]
Solving the above equation for x
$ \Rightarrow x = - \dfrac{1}{2}, - 1$
\[\mathop { \Rightarrow \lim }\limits_{a \to {0^ + }} \alpha \left( a \right)\] and \[\mathop {\lim }\limits_{a \to {0^ + }} \beta \left( a \right)\] are $ - \dfrac{1}{2}$and $ - 1$

So, option (D) is the correct answer.

Note: In some questions, we can just put the value in the limit and solve. But in many situations we cannot do this because we end up with the mathematically meaningless expression $\dfrac{0}{0}$ which could be anything. It is not always possible to work out limits simply by looking for factors and simplifying. So, we need to add binomial expansion and Taylor/Maclaurin series to our list of methods for working out limits as shown in the above solution.