Question

\(\underset{x\rightarrow 0}{lim}[\frac{log|2+x|-log|2-x|}{tanx}] =\) .......

• A. \(2\)
• B. \(0\)
• C. \(-1\)
• D. \(1\)

Hint:

Both numerator and denominator tend to zero, so use series or L Hopital rule, or write the standard limits.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
At \(x = 0\), the numerator is \(\log 2 - \log 2 = 0\) and the denominator is \(\tan 0 = 0\). So the form is \(0/0\) and we can use L'Hopital rule.

Step 2: Detailed Explanation:
Differentiate the top and bottom (near 0, \(2 + x\) and \(2 - x\) are positive so the modulus can be dropped).
\[ \frac{d}{dx}\left[\log(2+x) - \log(2-x)\right] = \frac{1}{2+x} + \frac{1}{2-x} \]
\[ \frac{d}{dx}\tan x = \sec^2 x \]
\[ \lim_{x\to 0}\frac{\frac{1}{2+x}+\frac{1}{2-x}}{\sec^2 x} = \frac{\frac12 + \frac12}{1} = 1 \]

Step 3: Check the options.
The limit is 1, which is not 0, 2 or \(-1\).

Final Answer:
The limit equals 1, option (D). \[ \boxed{1} \]