Question

The value of $\lim_{x\rightarrow1}\frac{\int_{2 \log_{e}x}^{3\log_{e} x} e^{t} \, dt}{x-1}$ equals

• A. 0
• B. 1
• C. -1
• D. $\infty$

Hint:

When the integrand is simple like $e^t$, you can also evaluate the integral first: $[e^t]_{2 \ln x}^{3 \ln x} = e^{3 \ln x} - e^{2 \ln x} = x^3 - x^2$. Then the limit $\lim_{x \to 1} \frac{x^3 - x^2}{x-1} = \lim_{x \to 1} \frac{x^2(x-1)}{x-1} = 1$.

Answer 1

The Correct Option is B

To find the value of this limit, we recognize that it represents an indeterminate form involving an integral function.
Step 1: As $x \to 1$, we substitute the value into the limits of the integral: Upper limit: $3 \log_e(1) = 0$. Lower limit: $2 \log_e(1) = 0$. The numerator becomes $\int_{0}^{0} e^t dt$, which is fundamentally 0. The denominator is $x - 1 = 1 - 1 = 0$. Since we have a $\frac{0}{0}$ form, we must apply L'Hôpital's Rule.
Step 2: The Leibniz Rule states: $\frac{d}{dx} \int_{g(x)}^{h(x)} f(t) dt = f(h(x))h'(x) - f(g(x))g'(x)$. Let $f(t) = e^t$, $h(x) = 3 \log_e x$, and $g(x) = 2 \log_e x$. Numerator derivative: $e^{3 \log_e x} \cdot \frac{d}{dx}(3 \log_e x) - e^{2 \log_e x} \cdot \frac{d}{dx}(2 \log_e x)$ $= e^{\log_e x^3} \cdot \frac{3}{x} - e^{\log_e x^2} \cdot \frac{2}{x}$ Using the identity $e^{\log_e u} = u$: $= x^3 \cdot \frac{3}{x} - x^2 \cdot \frac{2}{x} = 3x^2 - 2x$.
Step 3: The derivative of the denominator with respect to $x$ is straightforward: $\frac{d}{dx}(x - 1) = 1$.
Step 4: We now take the limit of the ratio of the derivatives: $\lim_{x \to 1} \frac{3x^2 - 2x}{1} = 3(1)^2 - 2(1) = 3 - 2 = 1$. The result is 1, which matches Option (2).