Question

If \(f(x) = \frac{3^x+3^{-x}-2}{tanx\cdot log(1+x)}\) for \(x\neq 0\), is continuous at \(x = 0\), then the value of \(f(0)\) is equal to \(\ldots\)

• A. \(2log3\)
• B. \((log3)^2\)
• C. \(\frac{1}{2}log3\)
• D. \(log\frac{1}{3}\)

Hint:

Write the numerator as a perfect square and use the standard limits for (a^x - 1)/x and log(1+x)/x.

Answer 1

The Correct Option is B

Step 1: Expand using series:
Let $L = \ln 3$. For small $x$: $3^x = 1 + Lx + \frac{L^2x^2}{2}$ and $3^{-x} = 1 - Lx + \frac{L^2x^2}{2}$.
So $3^x + 3^{-x} - 2 \approx L^2x^2$.

Step 2: Denominator:
$\tan x \approx x$ and $\ln(1+x) \approx x$, so the denominator is about $x^2$.

Step 3: Take the ratio:
$f(x) \to \dfrac{L^2x^2}{x^2} = L^2 = (\ln 3)^2$.

Step 4: Continuity:
Continuity at 0 means $f(0)$ equals this limit.

Final Answer:
$(\log 3)^2$, option (B). \[ \boxed{(\log 3)^2 \text{ (B)}} \]