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Limits, Continuity and Differentiability Part 3

Section: Mathematics

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Question : 34 of 79

Marks: +1, -0

For

\alpha, \beta, \gamma \in \mathbb{R}

\lim\limits_{x \to 0}\;\frac{x^2 \sin \alpha x + (\gamma - 1) e^{x^2}}{\sin 2x - \beta x} = 3

\beta + \gamma - \alpha

[2 Apr 2025 Shift 1]

4
6
7
-1

Solution:

\;\text{ At }\; x \to 0

\; \sin 2x - \beta x \to 0

\; \Rightarrow \;\; \frac{0}{0} \;\text{ form }\;

\; \Rightarrow \;\; (\gamma-1) e^0 + 0 \sin (\alpha x) \to 0

\; \Rightarrow \;\; (\gamma-1)=0

\; \Rightarrow \gamma=1

\; \Rightarrow \;\; \lim\limits_{x \to 0}\;\frac{x^2 \sin (\alpha x)}{\sin 2x - \beta x} = 3

\; \Rightarrow \;\; \lim\limits_{x \to 0}\;\frac{x^2 \left[ \alpha x - \frac{(\alpha x)^3}{3!} + \frac{(\alpha x)^5}{5!} - \cdots \right]}{\left[ (2x) - \frac{(2x)^3}{3!} + \frac{(2x)^5}{5!} - \cdots \right] - \beta x}

\; \Rightarrow \;\; \lim\limits_{x \to 0}\;\frac{\alpha x^3 - \frac{\alpha^3 x^5}{3!} + \frac{\alpha^5 x^7}{5!} - \cdots}{x(2-\beta) - \frac{8 x^3}{6} + \frac{2^5 \cdot x^5}{5!} - \cdots} = 3

\; \Rightarrow 2-\beta=0 \;\text{ and }\; \frac{\alpha}{\frac{-8}{6}} = 3

\; \Rightarrow \beta=2

\;\alpha = 3 \left( -\frac{8}{6} \right) = -4

\Rightarrow \gamma=1, \beta=2, \alpha=-4

\Rightarrow \beta+\gamma-\alpha=7

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