We are asked to evaluate the limit:

$ \lim_{x\rightarrow1}\frac{\int_{2\log_e x}^{3\log_e x}e^t \mathrm{dt}}{x-1} $

Indeterminate Form Check

As $x \rightarrow 1$, we have $\log_e x \rightarrow \log_e 1 = 0$. Therefore, the limits of the integral both approach $0$. The integral becomes $\int_{0}^{0}e^t \mathrm{dt} = 0$. The denominator $x-1$ also approaches $1-1 = 0$. This gives the indeterminate form $\frac{0}{0}$, which suggests using L'Hôpital's Rule.

Applying L'Hôpital's Rule

Let $f(x) = \int_{2\log_e x}^{3\log_e x}e^t \mathrm{dt}$ and $g(x) = x-1$. We need to find the limit of $\frac{f'(x)}{g'(x)}$.

1. Derivative of the denominator: $ g'(x) = \frac{d}{dx}(x-1) = 1 $
2. Derivative of the numerator: Using the Fundamental Theorem of Calculus and the chain rule: $ f'(x) = \frac{d}{dx}\left(\int_{2\log_e x}^{3\log_e x}e^t \mathrm{dt}\right) $ $ f'(x) = e^{3\log_e x} \cdot \frac{d}{dx}(3\log_e x) - e^{2\log_e x} \cdot \frac{d}{dx}(2\log_e x) $ Simplify the terms: $ e^{3\log_e x} = e^{\log_e x^3} = x^3 $ $ e^{2\log_e x} = e^{\log_e x^2} = x^2 $ $ \frac{d}{dx}(3\log_e x) = \frac{3}{x} $ $ \frac{d}{dx}(2\log_e x) = \frac{2}{x} $ Substitute back: $ f'(x) = x^3 \cdot \frac{3}{x} - x^2 \cdot \frac{2}{x} $ $ f'(x) = 3x^2 - 2x $
3. Evaluate the limit of the ratio of derivatives: $ \lim_{x\rightarrow1}\frac{f'(x)}{g'(x)} = \lim_{x\rightarrow1}\frac{3x^2 - 2x}{1} $ Substitute $x=1$: $ = 3(1)^2 - 2(1) = 3 - 2 = 1 $

The value of the limit is 1.