If lim_{t→0} (∫_{0}^{1} (3x+5)^t dx)^{1/t} = α/(5e) * (8/5)^{2/3} then α is equal to

Q: Wait, why do we use the $e^{\lim g(x)[f(x)-1]}$ formula? Can't we just take the natural log of both sides?

You absolutely can! Taking the natural log of the limit is the fundamental derivation of that formula. If $L = \lim [f(t)]^{g(t)}$, then $\ln L = \lim g(t) \ln(f(t))$. Since $f(t) \rightarrow 1$, $\ln(f(t)) \approx f(t)-1$ for small $t$. The formula is just a shortcut to save you time in the exam hall. Both methods are mathematically rigorous.

Q: Why can't we just differentiate the integral inside the limit directly using Leibniz's Rule?

You could, but it's often more complex. Leibniz's Rule for differentiation under the integral sign is powerful, but here, the integral is simple enough to evaluate directly. Evaluating it first gives you a clean algebraic expression, which is much easier to differentiate using L'Hopital's Rule than dealing with the derivative of the integral itself.

Q: How do I differentiate $8^{t+1}$ with respect to $t$?

Remember the rule for differentiating exponential functions: $\frac{d}{dt}(a^{u(t)}) = a^{u(t)} \ln(a) \cdot \frac{du}{dt}$. Here, $a=8$ and $u(t) = t+1$. So, the derivative is $8^{t+1} \ln(8) \cdot \frac{d}{dt}(t+1) = 8^{t+1} \ln(8) \cdot 1$. It's a standard chain rule application!

Question

If $\lim_{t\rightarrow0}(\int_{0}^{1}(3x+5)^{t}dx)^{\frac{1}{t}}=\frac{\alpha}{5e}(\frac{8}{5})^{\frac{2}{3}}$ then $\alpha$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the JEE Advanced landscape. Today, we are going to dissect a problem that looks like a monster but is actually a beautifully orchestrated dance of calculus.

We are tasked with finding the limit:
$$L = \lim_{t \rightarrow 0} \left( \int_{0}^{1} (3x+5)^t dx \right)^{\frac{1}{t}}$$

At first glance, this looks intimidating. An integral raised to a variable power requires a systematic approach.

### Identifying the Beast

In the world of limits, the first step is always to check the form. As $t \rightarrow 0$, the integrand $(3x+5)^t$ becomes $(3x+5)^0 = 1$. The integral of $1$ from $0$ to $1$ is simply $1$, while the exponent $\frac{1}{t}$ approaches infinity.

We have arrived at the classic $1^{\infty}$ indeterminate form. This is a signal to use the powerful identity:
$$\lim_{t \rightarrow 0} [f(t)]^{g(t)} = e^{\lim_{t \rightarrow 0} g(t)[f(t)-1]}$$

Our problem is now transformed into calculating the exponent:
$$L = \exp \left( \lim_{t \rightarrow 0} \frac{1}{t} \left( \int_{0}^{1} (3x+5)^t dx - 1 \right) \right)$$

### The Calculus of the Integral

Now, let's focus on the heart of the expression: the integral $I = \int_{0}^{1} (3x+5)^t dx$. Since we are integrating with respect to $x$, $t$ is treated as a constant.

We use the power rule for integration:
$$\int (ax+b)^n dx = \frac{(ax+b)^{n+1}}{a(n+1)}$$

Applying this, we get:
$$I = \left[ \frac{(3x+5)^{t+1}}{3(t+1)} \right]_{0}^{1}$$

Substituting the limits, we find:
$$I = \frac{8^{t+1} - 5^{t+1}}{3(t+1)}$$

### The L'Hopital Showdown

Now, we substitute $I$ back into our limit expression:
$$\lim_{t \rightarrow 0} \frac{1}{t} \left( \frac{8^{t+1} - 5^{t+1}}{3(t+1)} - 1 \right)$$

By taking the common denominator, we get:
$$\lim_{t \rightarrow 0} \frac{8^{t+1} - 5^{t+1} - 3(t+1)}{3t(t+1)}$$

Plugging in $t=0$ yields the $0/0$ form. Applying L'Hopital's Rule by differentiating the numerator and denominator with respect to $t$:
$$\lim_{t \rightarrow 0} \frac{8^{t+1} \ln(8) - 5^{t+1} \ln(5) - 3}{6t + 3}$$

Evaluating at $t=0$, we obtain:
$$\frac{8 \ln(8) - 5 \ln(5) - 3}{3}$$

### The Logarithmic Victory

We are almost there. We have the exponent value:
$$\frac{1}{3} (8 \ln(8) - 5 \ln(5) - 3)$$

Using logarithmic properties, this simplifies to:
$$\ln \left( \frac{8^{8/3}}{5^{5/3} e} \right)$$

Since this was the exponent of $e$, our final limit is:
$$L = \frac{8^{8/3}}{5^{5/3} e}$$

Breaking down the powers, $8^{8/3} = 64 \cdot 8^{2/3}$ and $5^{5/3} = 5 \cdot 5^{2/3}$. Thus, the final result is:
**$$L = \frac{64}{5e} \left(\frac{8}{5}\right)^{2/3}$$**

Animated Solution for Mathematics - Limits, Continuity and Differentiability: If $lim_{t \to 0} (\int_{0}^{1} (3 x + 5)^{t} d x)^{\frac{1}{t}} = \frac{α}{5 e} (\frac{8}{5})^{\frac{2}{3}}$ then $α$ is equal to

Enter Numerical Value:

Visualized Solution

Identifying the Limit Form

Applying the $1^{\infty}$ Formula

Evaluating the Definite Integral

Substituting the Limits of Integration

Setting up the Limit Expression

Applying L'Hopital's Rule

Evaluating the Limit at $t = 0$

Simplifying the Logarithmic Expression

Finding the Final Value of L

Comparing and Finding $α$

The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

Analyzing the Setup

Identifying the Beast

The Calculus of the Integral

The L'Hopital Showdown

The Logarithmic Victory

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Animated Solution for Mathematics - Limits, Continuity and Differentiability: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If limt→0​(∫01​(3x+5)tdx)t1​=5eα​(58​)32​ then α is equal to