Limits Question 4
Question 4 - 2024 (27 Jan Shift 2)
If $\lim _{x \rightarrow 0} \frac{3+\alpha \sin x+\beta \cos x+\log _{e}(1-x)}{3 \tan ^{2} x}=\frac{1}{3}$, then $2 \alpha-\beta$ is equal to :
(1) 2
(2) 7
(3) 5
(4) 1
Show Answer
Answer (3)
Solution
$$ \begin{aligned} & \lim _{x \rightarrow 0} \frac{3+\alpha \sin x+\beta \cos x+\log _{e}(1-x)}{3 \tan ^{2} x}=\frac{1}{3} \ & \Rightarrow \lim _{x \rightarrow 0} \frac{3+\alpha\left[x-\frac{x^{3}}{3 !}+\ldots\right]+\beta\left[1-\frac{x^{2}}{2 !}+\frac{x^{4}}{4 !} \ldots\right]+\left(-x-\frac{x^{2}}{2}-\frac{x^{3}}{3} \ldots\right)}{3 \tan ^{2} x}=\frac{1}{3} \ & \Rightarrow \lim _{x \rightarrow 0} \frac{(3+\beta)+(\alpha-1) x+\left(-\frac{1}{2}-\frac{\beta}{2}\right) x^{2}+\ldots}{3 x^{2}} \times \frac{x^{2} \text { natho }}{\tan ^{2} x}=\frac{1}{3} \ & \Rightarrow \beta+3=0, \alpha-1=0 \text { and } \frac{-\frac{1}{2}-\frac{\beta}{2}}{3}=\frac{1}{3} \ & \Rightarrow \beta=-3, \alpha=1 \ & \Rightarrow 2 \alpha-\beta=2+3=5 \end{aligned} $$
