Limits & Continuity
Two limits; finding 1/(4Lā-m/n)
MMTS_Full_Test_04
Grade 12
Question:
Let $L_1=\displaystyle\lim_{x\to0}\frac{\tan3x-\tan x}{2\cos4x\sin3x+\sin5x-(2-2\cos4x+\sin3x)}$ and $L_2=\displaystyle\lim_{x\to0^+}\frac{e^{\sec(\tan x^{1012n})}-e}{\sin(x^{2025m})}=\frac{e}{2}$; $n,m\in\mathbb{N}$. Then $\dfrac{1}{4L_1-\frac{m}{n}}$ is
(A) 1012
(B) 2024
(C) 2025
(D) $\dfrac{1}{2025}$
Step-by-Step Solution
Key Concept: $L_1=1/4$ (expand to leading order). $L_2=e/2$ gives $2024n=2025m\Rightarrow m/n=2024/2025$.
$\dfrac{1}{4L_1-m/n}=2025$.
Correct Answer: (C) 2025