If lim (x→0) [1 + x ln(1 + b²)]^(1/x) = 2b sin² θ, b 0 and θ ∈ (-π, π], then the value of θ is (a)

Q: Why can't I just use L'Hopital's Rule for the limit?

You absolutely can! But L'Hopital on $1^{\infty}$ requires taking the log first, which leads to $\frac{\ln(f(x))}{1/g(x)}$. The standard formula $e^{\lim (f-1)g}$ is essentially a shortcut for that exact process. It saves time and reduces the chance of calculation errors.

Q: How did you know to use AM-GM here?

It's a pattern recognition skill. Whenever you see an expression of the form $x + \frac{1}{x}$ where $x > 0$, your brain should immediately scream 'AM-GM Inequality!' It's the most powerful tool for finding the minimum value of such expressions.

Q: Why is $\sin^2 \theta = 1$ the only solution?

Because we found two conflicting bounds. Algebraically, we proved $\sin^2 \theta \ge 1$. Trigonometrically, we know $\sin^2 \theta \le 1$. The only way for both to be true is if $\sin^2 \theta = 1$. It's like a mathematical pincer movement!

Question

If $\lim_{x 	o 0} [1 + x \ln(1 + b^2)]^{1/x} = 2b \sin^2 	heta, b > 0$ and $	heta \in (-\pi, \pi]$, then the value of $	heta$ is

Options:

(A) $\pm \frac{\pi}{4}$

(B) $\pm \frac{\pi}{3}$

(C) $\pm \frac{\pi}{6}$

(D) $\pm \frac{\pi}{2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The problem presents the limit:
$$\lim_{x 	o 0} [1 + x \ln(1 + b^2)]^{1/x} = 2b \sin^2 	heta$$

As $x 	o 0$, the base approaches $1$ and the exponent approaches $\infty$. This is the classic $1^{\infty}$ indeterminate form.

To resolve this, we utilize the standard limit identity:
$$\lim_{x 	o a} [f(x)]^{g(x)} = e^{\lim_{x 	o a} (f(x)-1)g(x)}$$

### The Algebraic Bridge

We identify $f(x) = 1 + x \ln(1 + b^2)$ and $g(x) = 1/x$. Subtracting $1$ from $f(x)$ yields $x \ln(1 + b^2)$.

Multiplying this by $g(x) = 1/x$ results in the cancellation of $x$:
$$(f(x)-1)g(x) = x \ln(1 + b^2) \cdot \frac{1}{x} = \ln(1 + b^2)$$

Applying the exponential function, the limit simplifies to:
$$e^{\ln(1 + b^2)} = 1 + b^2$$

### The AM-GM Insight

Equating our result to the right-hand side of the original equation, we have:
$$1 + b^2 = 2b \sin^2 	heta$$

Isolating $\sin^2 	heta$, we obtain:
$$\sin^2 	heta = \frac{1 + b^2}{2b} = \frac{1}{2} \left( b + \frac{1}{b} ight)$$

Given $b > 0$, we apply the Arithmetic Mean-Geometric Mean (AM-GM) inequality:
$$\frac{b + \frac{1}{b}}{2} \ge \sqrt{b \cdot \frac{1}{b}} = 1$$

This implies that $\sin^2 	heta \ge 1$.

### The Final Pincer Movement

We must reconcile our result $\sin^2 	heta \ge 1$ with the fundamental trigonometric constraint $0 \le \sin^2 	heta \le 1$.

These conditions can only be satisfied simultaneously if:
$$\sin^2 	heta = 1$$

This implies $\sin 	heta = 1$ or $\sin 	heta = -1$. Within the interval $	heta \in (-\pi, \pi]$, the solutions are:
$$	heta = \frac{\pi}{2}, -\frac{\pi}{2}$$

Thus, the final values for $	heta$ are **$\pm \frac{\pi}{2}$**.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: If $lim_{x \to 0} [1 + x ln (1 + b^{2})]^{1/ x} = 2 b sin^{2} θ, b > 0$ and $θ \in (- π, π]$ , then the value of $θ$ is

Select Answer:

Visualized Solution

Analyze the Limit Form

Standard Limit Formula

Substitute into Formula

Simplify the Exponent

Final Limit Value

Set Up the Equation

Isolate $sin^{2} θ$

AM-GM Inequality

Constraint Check

Visualizing the Bounds

Deducing $sin^{2} θ = 1$

Final Solution

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

Analyzing the Setup

The Algebraic Bridge

The AM-GM Insight

The Final Pincer Movement

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