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: Combine the logs.
\[ \log(2+x) - \log(2-x) = \log\frac{1 + x/2}{1 - x/2} \]

Step 2: Expand.
Using $\log(1 + u) \approx u$ for small $u$: $\log(1 + x/2) - \log(1 - x/2) \approx \dfrac{x}{2} + \dfrac{x}{2} = x$.

Step 3: Divide by tan x.
For small $x$, $\tan x \approx x$, so the ratio tends to $x / x = 1$.

Final Answer:
Option (D). \[ \boxed{1} \]