Question

If $f(x)$ is a polynomial of least degree, such that $\lim_{x\to0}\left(1+\frac{f(x)+x^2}{x^2}\right)^{1/x}=e^2$, then $f(2)$ is -

• A. 2
• B. 8
• C. 10
• D. 12

Answer 1

Answer: (D)

12

Answer 2

The given limit is $\lim_{x\to0}\left(1+\frac{f(x)+x^2}{x^2}\right)^{1/x}=e^2$. This limit is of the form $1^\infty$. We use the property $\lim_{x\to a} (1+g(x))^{h(x)} = e^{\lim_{x\to a} g(x)h(x)}$. Here, $g(x) = \frac{f(x)+x^2}{x^2}$ and $h(x) = \frac{1}{x}$. For the limit to be finite, $\lim_{x\to0} g(x) = \lim_{x\to0} \frac{f(x)+x^2}{x^2} = 0$. This implies that $f(x)$ must have terms that cancel out the $x^2$ in the denominator and result in terms of degree higher than 2. Thus, $f(x)$ must be of the form $f(x) = -x^2 + a_3 x^3 + a_4 x^4 + \dots$. Now, consider the exponent: $\lim_{x\to0} g(x)h(x) = \lim_{x\to0} \left(\frac{f(x)+x^2}{x^2}\right) \left(\frac{1}{x}\right)$ Substitute $f(x) = -x^2 + a_3 x^3 + a_4 x^4 + \dots$: $\lim_{x\to0} \left(\frac{-x^2 + a_3 x^3 + a_4 x^4 + \dots + x^2}{x^2}\right) \left(\frac{1}{x}\right)$ $= \lim_{x\to0} \left(\frac{a_3 x^3 + a_4 x^4 + \dots}{x^2}\right) \left(\frac{1}{x}\right)$ $= \lim_{x\to0} (a_3 x + a_4 x^2 + \dots) \left(\frac{1}{x}\right)$ $= \lim_{x\to0} (a_3 + a_4 x + \dots)$ This limit equals $a_3$. We are given that the original limit is $e^2$, so the exponent must be 2: $a_3 = 2$. Since $f(x)$ is a polynomial of the least degree, we set all higher-order coefficients to zero. Thus, $f(x) = -x^2 + 2x^3$. We need to find $f(2)$: $f(2) = -(2)^2 + 2(2)^3 = -4 + 2(8) = -4 + 16 = 12$.