Question

$\lim_{x\to0} \frac{e^{-(1+2x)^{\frac{1}{2x}}}}{x}$ is equal to:

• A. e
• B. $\frac{-2}{e}$
• C. 0
• D. $e-e^2$

Answer 1

The question as stated does not yield a finite answer among the options. There is likely a typo in the question.

Answer 2

Let the limit be $L$. We first evaluate the exponent term $(1+2x)^{\frac{1}{2x}}$. This is of the indeterminate form $1^\infty$. Let $y = (1+2x)^{\frac{1}{2x}}$. Then $\ln y = \frac{1}{2x} \ln(1+2x)$. Using the standard limit $\lim_{u\to0} \frac{\ln(1+u)}{u} = 1$, or by Taylor series expansion of $\ln(1+u) = u - \frac{u^2}{2} + \dots$, we get: $\ln y = \frac{1}{2x} \left( 2x - \frac{(2x)^2}{2} + \dots \right) = \frac{1}{2x} \left( 2x - 2x^2 + \dots \right) = 1 - x + \dots$. So, $y = e^{1 - x + \dots}$. As $x \to 0$, $y \to e^1 = e$. The expression in the limit is $\frac{e^{-y}}{x}$. As $x \to 0$, $y \to e$. So the numerator $e^{-y} \to e^{-e}$. The limit becomes $\lim_{x\to0} \frac{e^{-e}}{x}$. Since $e^{-e}$ is a non-zero constant, and the denominator approaches 0, the limit tends to $\pm \infty$. This indicates that the question as stated might have a typo or the options are for a different question.