If \(\lim_{{x \to 0}} \frac{\alpha e^x + \beta e^{-x} + \gamma \sin x}{x \sin^2 x} = \frac{2}{3}\), where \(\alpha, \beta, \gamma \in \mathbb{R}\) then which of the following is NOT correct?
\(\alpha^2 + \beta^2 + \gamma^2 = 6\)
\(\alpha \beta + \beta \gamma + \gamma \alpha + 1 = 0\)
\(\alpha \beta^2 + \beta \gamma^2 + \gamma \alpha^2 + 3 = 0\)
\(\alpha^2 - \beta^2 + \gamma^2 = 4\)
The Correct Option is C
Solution and Explanation
\(\lim_{{x \to 0}} \frac{\alpha e^x + \beta e^{-x} + \gamma \sin x}{x \sin^2 x} = \frac{2}{3}\)
\(⇒ α + β = 0\) (to make indeterminant form) …(i)
Now,
\(\lim_{{x \to 0}} \frac{\alpha e^x + \beta e^{-x} + \gamma \sin x}{x \sin^2 x} = \frac{2}{3}\)
(Using L-H Rule)
\(⇒ α – β + γ = 0\) (to make indeterminant form) …(ii)
Now,
\(\lim_{{x \to 0}} \frac{\alpha e^x + \beta e^{-x} + \gamma \sin x}{6x} = \frac{2}{3}\)
(Using L-H Rule)
\(⇒\)\(\frac{\alpha - \beta - \gamma}{6} = \frac{2}{3}\)
\(⇒\)\(α – β – γ = 4 …(iii)\)
\(⇒\) \(γ = –2\)
and eq(i) + eq(ii)
\(2α = –γ\)
On solving,
\(⇒ α = 1\ \text{and}\ β = –1\)
and \(\alpha \beta^2 + \beta \gamma^2 + \gamma \alpha^2 + 3\)
\(= 1 – 4 – 2 + 3\)
\(= –2\)
So, the correct option is (C): \(\alpha \beta^2 + \beta \gamma^2 + \gamma \alpha^2 + 3 = 0\)
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Concepts Used:
Limits
A function's limit is a number that a function reaches when its independent variable comes to a certain value. The value (say a) to which the function f(x) approaches casually as the independent variable x approaches casually a given value "A" denoted as f(x) = A.
If limx→a- f(x) is the expected value of f when x = a, given the values of ‘f’ near x to the left of ‘a’. This value is also called the left-hand limit of ‘f’ at a.
If limx→a+ f(x) is the expected value of f when x = a, given the values of ‘f’ near x to the right of ‘a’. This value is also called the right-hand limit of f(x) at a.
If the right-hand and left-hand limits concur, then it is referred to as a common value as the limit of f(x) at x = a and denote it by lim x→a f(x).
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