Question

If $\lim_{x\to 0} (1+ax+bx^2)^{2/x} = e^e$, then

• A. a=3/2 and b in R
• B. a=3/2 and b in R+
• C. a=0 and b=1
• D. a=1 and b=0

Answer 1

The provided options do not match the exact question. If the question intended the limit to be $e^3$, then option (A) would be correct. Based on the exact question $\lim_{x\to 0} (1+ax+bx^2)^{2/x} = e^e$, we find $a = e/2$ and $b \in \mathbb{R}$. Since this is not an option, and assuming there's a typo in the question and it should be $e^3$, then $a=3/2$ and $b \in \mathbb{R}$ is the intended answer.

Answer 2

The limit is of the indeterminate form $1^\infty$. We can evaluate it using the formula $\lim_{x\to c} (f(x))^{g(x)} = e^{\lim_{x\to c} g(x)(f(x)-1)}$ or by using Taylor series expansion.

Let $L = \lim_{x\to 0} (1+ax+bx^2)^{2/x}$. We can write $L$ as: $L = \exp\left(\lim_{x\to 0} \frac{2}{x} \ln(1+ax+bx^2)\right)$

Now, we evaluate the limit in the exponent: $\lim_{x\to 0} \frac{2}{x} \ln(1+ax+bx^2)$ As $x \to 0$, $ax+bx^2 \to 0$. We use the Taylor expansion of $\ln(1+u)$ around $u=0$, which is $\ln(1+u) = u - \frac{u^2}{2} + O(u^3)$. Let $u = ax+bx^2$. $\ln(1+ax+bx^2) = (ax+bx^2) - \frac{(ax+bx^2)^2}{2} + O((ax+bx^2)^3)$ $\ln(1+ax+bx^2) = ax+bx^2 - \frac{a^2x^2 + 2abx^3 + b^2x^4}{2} + O(x^3)$ $\ln(1+ax+bx^2) = ax + \left(b-\frac{a^2}{2}\right)x^2 + O(x^3)$ Substitute this back into the exponent's limit: $\lim_{x\to 0} \frac{2}{x} \left( ax + \left(b-\frac{a^2}{2}\right)x^2 + O(x^3) \right)$ $= \lim_{x\to 0} 2 \left( a + \left(b-\frac{a^2}{2}\right)x + O(x^2) \right)$ $= 2(a + 0) = 2a$ So, the limit $L$ is $e^{2a}$.

The problem states that $L = e^e$. $e^{2a} = e^e$ Equating the exponents, we get: $2a = e \implies a = \frac{e}{2}$ The value of $b$ does not affect the limit as $x \to 0$, because the term involving $b$ in the Taylor expansion of the logarithm is multiplied by $x^2$, and then by $2/x$, resulting in a term proportional to $x$, which vanishes as $x \to 0$. Thus, $b$ can be any real number ($b \in \mathbb{R}$).

The derived values are $a=e/2$ and $b \in \mathbb{R}$. However, $a=e/2$ is not present in any of the given options. This strongly suggests a typo in the question, where $e^e$ was likely intended to be $e^3$. This is a common type of typo in limit problems of this nature.

If we assume the question intended the limit to be $e^3$: If $\lim_{x\to 0} (1+ax+bx^2)^{2/x} = e^3$, then: $e^{2a} = e^3$ Equating the exponents: $2a = 3 \implies a = \frac{3}{2}$ As established before, the value of $b$ does not affect the limit. Therefore, $b$ can be any real number ($b \in \mathbb{R}$).

Under this assumption, we have $a=3/2$ and $b \in \mathbb{R}$. This matches option (A).