Let α be a positive real number. Let f: ℝ → ℝ and g: (α, ∞) → ℝ be the functions defined by f(x) = s

Q: Why are we allowed to just push the limit inside the sine function?

Because $f(x) = \sin(\frac{\pi x}{12})$ is a continuous function. For any continuous function $f$, the limit of the composition is the composition of the limits: $\lim_{x \to a} f(g(x)) = f(\lim_{x \to a} g(x))$. This is a massive time-saver that turns a complex problem into a simpler one!

Q: I got stuck at the $\frac{\infty}{\infty}$ form. How do I know if I should use L'Hopital's or something else?

L'Hopital's Rule is your best friend for $\frac{0}{0}$ or $\frac{\infty}{\infty}$ indeterminate forms. Since our expression involves logarithms and exponentials, algebraic manipulation would be a nightmare. L'Hopital's is the most direct path to the solution.

Q: Is it okay to use L'Hopital's Rule twice in the same problem?

Absolutely! As long as the resulting expression is still in an indeterminate form like $\frac{0}{0}$ or $\frac{\infty}{\infty}$, you can apply the rule again. It is like peeling an onion—sometimes you need to peel more than one layer to get to the core.

Question

Let $\alpha$ be a positive real number. Let $f: \mathbb{R} ightarrow \mathbb{R}$ and $g: (\alpha, \infty) ightarrow \mathbb{R}$ be the functions defined by $f(x) = \sin \left( \frac{\pi x}{12} ight)$ and $g(x) = \frac{2 \log_e (\sqrt{x} - \sqrt{\alpha})}{\log_e (e^{\sqrt{x}} - e^{\sqrt{\alpha}})}$. Then the value of $\lim_{x 	o \alpha^+} f(g(x))$ is _______.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are tasked with finding the value of $\lim_{x 	o \alpha^+} f(g(x))$, where $f(x) = \sin(\frac{\pi x}{12})$ and $g(x) = \frac{2 \log_e (\sqrt{x} - \sqrt{\alpha})}{\log_e (e^{\sqrt{x}} - e^{\sqrt{\alpha}})}$.

At first glance, this looks like a labyrinth of logarithms and exponentials. However, the secret to mastering JEE Advanced problems is not brute force; it is elegance.

### The Continuity Shortcut

The first step is to look at the outer function, $f(x) = \sin(\frac{\pi x}{12})$. Since sine functions are continuous everywhere, we can invoke the property that:

$$\lim_{x 	o \alpha^+} f(g(x)) = f\left(\lim_{x 	o \alpha^+} g(x)ight)$$

This allows us to focus entirely on the inner function $g(x)$ and find its limit, which we will call $L$. Once we have $L$, the final answer is simply $f(L)$.

We have effectively reduced the problem to finding:

$$L = \lim_{x 	o \alpha^+} \frac{2 \log_e (\sqrt{x} - \sqrt{\alpha})}{\log_e (e^{\sqrt{x}} - e^{\sqrt{\alpha}})}$$

### Confronting the Beast

Now, let us examine $g(x)$ as $x$ approaches $\alpha$ from the right. As $x 	o \alpha^+$, the term $\sqrt{x} - \sqrt{\alpha}$ approaches $0^+$.

Consequently, the numerator $2 \log_e (\sqrt{x} - \sqrt{\alpha})$ approaches $2 \log_e(0^+)$, which is $-\infty$. Similarly, in the denominator, $e^{\sqrt{x}} - e^{\sqrt{\alpha}}$ approaches $0^+$, so $\log_e (e^{\sqrt{x}} - e^{\sqrt{\alpha}})$ also approaches $-\infty$.

We have arrived at the classic $\frac{\infty}{\infty}$ indeterminate form. This is the perfect invitation to use **L'Hopital's Rule**.

### The L'Hopital Dance

We differentiate the numerator and the denominator separately. For the numerator, we use the chain rule:

$$\frac{d}{dx} [2 \log_e (\sqrt{x} - \sqrt{\alpha})] = 2 \cdot \frac{1}{\sqrt{x} - \sqrt{\alpha}} \cdot \frac{1}{2\sqrt{x}}$$

For the denominator, we also use the chain rule:

$$\frac{d}{dx} [\log_e (e^{\sqrt{x}} - e^{\sqrt{\alpha}})] = \frac{1}{e^{\sqrt{x}} - e^{\sqrt{\alpha}}} \cdot e^{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}}$$

When we put these back into our limit, we see a beautiful cancellation: the term $\frac{1}{2\sqrt{x}}$ appears in both the numerator and the denominator. They vanish, leaving us with:

$$L = \lim_{x 	o \alpha^+} \frac{2(e^{\sqrt{x}} - e^{\sqrt{\alpha}})}{e^{\sqrt{x}}(\sqrt{x} - \sqrt{\alpha})}$$

### The Second Act

We can split this limit into two parts:

$$L = \lim_{x 	o \alpha^+} \left(\frac{2}{e^{\sqrt{x}}}ight) \cdot \lim_{x 	o \alpha^+} \left(\frac{e^{\sqrt{x}} - e^{\sqrt{\alpha}}}{\sqrt{x} - \sqrt{\alpha}}ight)$$

The first part is simple: as $x 	o \alpha$, it becomes $\frac{2}{e^{\sqrt{\alpha}}}$. The second part is another $\frac{0}{0}$ form.

Applying L'Hopital's Rule again, we differentiate the numerator $e^{\sqrt{x}} - e^{\sqrt{\alpha}}$ to get $e^{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}}$, and the denominator $\sqrt{x} - \sqrt{\alpha}$ to get $\frac{1}{2\sqrt{x}}$. Again, the $\frac{1}{2\sqrt{x}}$ terms cancel, leaving us with $\lim_{x 	o \alpha^+} e^{\sqrt{x}} = e^{\sqrt{\alpha}}$.

### The Grand Finale

Combining our results, we find:

$$L = \left(\frac{2}{e^{\sqrt{\alpha}}}ight) \cdot (e^{\sqrt{\alpha}}) = 2$$

All that complexity has collapsed into the integer $2$. Finally, we return to our original function:

$$f(L) = f(2) = \sin\left(\frac{\pi \cdot 2}{12}ight) = \sin\left(\frac{\pi}{6}ight) = \mathbf{0.5}$$

We have conquered the beast! Remember, in mathematics, persistence and a calm mind are your greatest tools.

Enter Numerical Value:

Visualized Solution

Limit of a Composite Function

Setting up $g (x)$

Identifying the Indeterminate Form

Applying L'Hopital's Rule

Differentiating the Numerator

Differentiating the Denominator

Substituting the Derivatives

Simplifying the Expression

Re-evaluating the Limit Form

Second Application of L'Hopital's Rule

Evaluating the Second Limit

Combining the Results for $L$

Final Substitution into $f (x)$

Final Answer

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

Analyzing the Setup

The Continuity Shortcut

Confronting the Beast

The L'Hopital Dance

The Second Act

The Grand Finale

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