Question

If $\displaystyle \lim_{x \to 0} \frac{(1+3x+2x^2)^{1/x}-(1+3x-2x^2)^{1/x}}{x}=2ke^3$ then the value of $k$ is

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (B)

2

Answer 2

The limit is evaluated using the Taylor expansion of the form $(1+ax+bx^2)^{1/x} = e^a(1+(b-a^2/2)x + O(x^2))$. Applying this to both terms in the numerator with $a=3$ and $b=2$ for the first term and $b=-2$ for the second term, we get $e^3(1 - 5x/2)$ and $e^3(1 - 13x/2)$ respectively. Substituting these into the limit expression and simplifying yields $4e^3$. Equating this to $2ke^3$ gives $k=2$.