Question:
What is \(\lim_{x \to 0} \frac{\tan x}{\sin 2x}\) equal to?
\(\lim_{x \to 0} \frac{\tan x}{\sin 2x}\) किसके बराबर है?
Options:
- 1/2
- 2
- 1
- Limit does not exist
- 1/2
- 2
- 1
- सीमा मौजूद नहीं है
Correct Answer: a)
Explanation:
Using the standard limits \(\lim_{x \to 0} \frac{\tan x}{x} = 1\) and \(\lim_{x \to 0} \frac{\sin 2x}{2x} = 1\):
\[\lim_{x \to 0} \frac{\tan x}{\sin 2x} = \lim_{x \to 0} \frac{\frac{\tan x}{x}}{2 \cdot \frac{\sin 2x}{2x}} = \frac{1}{2 \times 1} = \frac{1}{2}\]
Alternatively, \(\frac{\tan x}{\sin 2x} = \frac{\sin x / \cos x}{2\sin x \cos x} = \frac{1}{2\cos^2 x} \to \frac{1}{2}\) as \(x \to 0\).
Hence, the correct option is (a).
मानक सीमाओं \(\lim_{x \to 0} \frac{\tan x}{x} = 1\) तथा \(\lim_{x \to 0} \frac{\sin 2x}{2x} = 1\) का उपयोग करने पर:
\[\lim_{x \to 0} \frac{\tan x}{\sin 2x} = \lim_{x \to 0} \frac{\tan x / x}{2 (\sin 2x / 2x)} = \frac{1}{2(1)} = \frac{1}{2}\]
अतः सही विकल्प (a) है।