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Question: Let L= \({\lim _{x \to 0}}\) \(\dfrac{{a - \sqrt {{a^2} - {x^2}} - \dfrac{{{x^2}}}{4}}}{{{x^4}}}\), ...

Let L= lim⁡x→0{\lim _{x \to 0}} a−a2−x2−x24x4\dfrac{{a - \sqrt {{a^2} - {x^2}} - \dfrac{{{x^2}}}{4}}}{{{x^4}}}, a>0a > 0. If L is finite, then
(THIS QUESTION HAS MULTIPLE CORRECT OPTIONS)
A. a=2
B. a=1
C. L=164\dfrac{1}{{64}}
D. L=132\dfrac{1}{{32}}

Explanation

Solution

To solve this question, we need to know the basic theory related to the chapter limits. As we know, when we put x→0x \to 0, we will get an indeterminate form. Thus, here we will manipulate the given expression in such a way that we will get its approaching value or limit value.

Complete step-by-step answer :
As given in the question,
lim⁡x→0{\lim _{x \to 0}} a−a2−x2−x24x4\dfrac{{a - \sqrt {{a^2} - {x^2}} - \dfrac{{{x^2}}}{4}}}{{{x^4}}}
It is to be noted that, on substituting the value x→0x \to 0 directly to the function, the numerator as well as denominator will become 0, and we know the value 00\dfrac{0}{0}, does not exist.
lim⁡x→0{\lim _{x \to 0}} a−a2−x2−x24x4\dfrac{{a - \sqrt {{a^2} - {x^2}} - \dfrac{{{x^2}}}{4}}}{{{x^4}}}
lim⁡x→0{\lim _{x \to 0}} a - a(1 - x2a2)12 - x24x4\dfrac{{{\text{a - a}}{{\left( {{\text{1 - }}\dfrac{{{{\text{x}}^{\text{2}}}}}{{{{\text{a}}^{\text{2}}}}}} \right)}^{\dfrac{{\text{1}}}{{\text{2}}}}}{\text{ - }}\dfrac{{{{\text{x}}^{\text{2}}}}}{{\text{4}}}}}{{{{\text{x}}^{\text{4}}}}}
Here, we use the expansion of (1+x)n{\left( {1 + x} \right)^n}.
As we know, (1+x)n{\left( {1 + x} \right)^n}= 1+nx+n(n−1)2!x2+......1 + nx + \dfrac{{n\left( {n - 1} \right)}}{{2!}}{x^2} + ......

lim⁡x→0{\lim _{x \to 0}} a - a(1 - 12x2a2 + 12(12 - 1)2( - x2a2)2) - x24x4\dfrac{{{\text{a - a}}\left( {{\text{1 - }}\dfrac{{\text{1}}}{{\text{2}}}\dfrac{{{{\text{x}}^{\text{2}}}}}{{{{\text{a}}^{\text{2}}}}}{\text{ + }}\dfrac{{\dfrac{{\text{1}}}{{\text{2}}}\left( {\dfrac{{\text{1}}}{{\text{2}}}{\text{ - 1}}} \right)}}{{\text{2}}}{{\left( {{\text{ - }}\dfrac{{{{\text{x}}^{\text{2}}}}}{{{{\text{a}}^{\text{2}}}}}} \right)}^{\text{2}}}} \right){\text{ - }}\dfrac{{{{\text{x}}^{\text{2}}}}}{{\text{4}}}}}{{{{\text{x}}^{\text{4}}}}}
lim⁡x→0{\lim _{x \to 0}} a - a(1 - 12x2a2 - 18x4a4) - x24x4\dfrac{{{\text{a - a}}\left( {{\text{1 - }}\dfrac{{\text{1}}}{{\text{2}}}\dfrac{{{{\text{x}}^{\text{2}}}}}{{{{\text{a}}^{\text{2}}}}}\dfrac{{{\text{ - 1}}}}{{\text{8}}}\dfrac{{{{\text{x}}^{\text{4}}}}}{{{{\text{a}}^{\text{4}}}}}} \right){\text{ - }}\dfrac{{{{\text{x}}^{\text{2}}}}}{{\text{4}}}}}{{{{\text{x}}^{\text{4}}}}}
lim⁡x→0{\lim _{x \to 0}} 12x2a + 18x4a3 - x24x4\dfrac{{\dfrac{{\text{1}}}{{\text{2}}}\dfrac{{{{\text{x}}^{\text{2}}}}}{{\text{a}}}{\text{ + }}\dfrac{{\text{1}}}{{\text{8}}}\dfrac{{{{\text{x}}^{\text{4}}}}}{{{{\text{a}}^{\text{3}}}}}{\text{ - }}\dfrac{{{{\text{x}}^{\text{2}}}}}{{\text{4}}}}}{{{{\text{x}}^{\text{4}}}}}
Limit is finite when
⇒\Rightarrow 12a=14\dfrac{1}{{2a}} = \dfrac{1}{4}
⇒\Rightarrow a = 2{\text{a = 2}}
∴L=18a3\therefore L = \dfrac{1}{{8{a^3}}}
∴L=18a3\therefore L = \dfrac{1}{{8{a^3}}}
∴L=164\therefore L = \dfrac{1}{{64}}
∴L\therefore L =164,a=2 = \dfrac{1}{{64}},a = 2
Therefore, option (A) and (C) are the correct answer.

Note : A function may approach two different limits. One where the variable approaches its limit through values larger than the limit and the other where the variable approaches its limit through values smaller than the limit. In such a case, the limit is not defined but the right and left-hand limit exist.