Question

\(\lim _{h\rightarrow 0}\frac{\log _{e}(1+2h)-2\log _{e}(1+h)}{h^{2}}\)

• A. \(-1\)
• B. \(\infty\)
• C. \(e\)
• D. \(\frac{1}{3}\)

Hint:

Expand both logarithms as series up to h squared.

Answer 1

The Correct Option is A

Step 1: Expand
\(\log_e(1+2h)=2h-\dfrac{(2h)^2}{2}+\cdots=2h-2h^2+\cdots\)

Step 2: Second term
\(2\log_e(1+h)=2\left(h-\dfrac{h^2}{2}+\cdots\right)=2h-h^2+\cdots\)

Step 3: Subtract
\(\log_e(1+2h)-2\log_e(1+h)=-h^2+\text{higher terms}\)

Step 4: Take the limit
\(\displaystyle\lim_{h\to0}\dfrac{-h^2+\cdots}{h^2}=-1\)

Final Answer:
The limit equals -1. \[ \boxed{\text{(A) }-1} \]