If α = lim(x→0+) (e^√tan x - e^√x) / (√tan x - √x) and β = lim(x→0) (1 + sin x)^(1/2 cot x) are the

Q: Wait, why do we factor out $e^{\sqrt{x}}$ in the limit for $\alpha$?

Great question! We factor out $e^{\sqrt{x}}$ to transform the expression into the form $e^{\sqrt{x}} \cdot \frac{e^{\sqrt{\tan x} - \sqrt{x}} - 1}{\sqrt{\tan x} - \sqrt{x}}$. This allows us to use the standard limit $\lim_{t \to 0} \frac{e^t-1}{t} = 1$, where $t = \sqrt{\tan x} - \sqrt{x}$. It turns a complex expression into a simple, recognizable one.

Q: Can we use L'Hopital's rule for $\beta$ instead of the $1^{\infty}$ formula?

You certainly could, but it would be significantly more tedious! Using L'Hopital's rule on $(1+\sin x)^{\frac{1}{2} \cot x}$ requires taking the logarithm first, which leads to a complex derivative. The $1^{\infty}$ formula $\lim f(x)^{g(x)} = e^{\lim (f(x)-1)g(x)}$ is a powerful, specialized tool designed exactly for this type of problem. It saves time and reduces the chance of calculation errors.

Q: Why did we multiply the quadratic equation by $-1$ at the end?

This is a classic trap! The problem gives us the equation $ax^2 + bx - \sqrt{e} = 0$. Our derived equation was $x^2 - (1+\sqrt{e})x + \sqrt{e} = 0$. Notice the constant term: ours is $+\sqrt{e}$, but the given one is $-\sqrt{e}$. To compare coefficients correctly, we must multiply our entire equation by $-1$ to make the constant terms match. Always check your signs when comparing coefficients!

Question

If $\alpha = \lim_{x 	o 0^+} \left( \frac{e^{\sqrt{	an x}} - e^{\sqrt{x}}}{\sqrt{	an x} - \sqrt{x}} ight)$ and $\beta = \lim_{x 	o 0} (1 + \sin x)^{\frac{1}{2} \cot x}$ are the roots of the quadratic equation $ax^2 + bx - \sqrt{e} = 0$, then $12 \log_e(a+b)$ is equal to__________

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the JEE journey. Today, we are not just solving a problem; we are conducting a symphony of mathematical concepts.

We have a limit that looks like a tangled knot and a quadratic equation that acts as our final destination. Let us unravel this together.

### Unmasking $\alpha$

We begin with the limit:
$$\alpha = \lim_{x 	o 0^+} \left( \frac{e^{\sqrt{	an x}} - e^{\sqrt{x}}}{\sqrt{	an x} - \sqrt{x}} ight)$$

At first glance, this looks intimidating. The square roots and the exponential functions seem to be fighting for dominance. But remember, in limits, structure is everything.

We want to reach the standard form $\lim_{t 	o 0} \frac{e^t-1}{t} = 1$. To do this, we perform a strategic maneuver: we factor out $e^{\sqrt{x}}$ from the numerator.

This gives us:
$$\alpha = \lim_{x 	o 0^+} e^{\sqrt{x}} \left( \frac{e^{\sqrt{	an x} - \sqrt{x}} - 1}{\sqrt{	an x} - \sqrt{x}} ight)$$

Now, look at the bracketed term. If we let $t = \sqrt{	an x} - \sqrt{x}$, then as $x 	o 0^+$, $t$ also approaches $0$. The expression inside the limit becomes $e^{\sqrt{x}} \cdot 1$.

Since $e^{\sqrt{0}} = 1$, we find that **$\alpha = 1$**. The knot is untangled.

### The $1^{\infty}$ Dance with $\beta$

Next, we face the limit:
$$\beta = \lim_{x 	o 0} (1 + \sin x)^{\frac{1}{2} \cot x}$$

If you try to plug in $x=0$ directly, you get $1^{\infty}$, a classic indeterminate form that JEE examiners love to test. We don't panic; we use our trusty formula:
$$\lim_{x 	o a} f(x)^{g(x)} = e^{\lim_{x 	o a} (f(x)-1)g(x)}$$

Applying this, we get:
$$\beta = e^{\lim_{x 	o 0} (1 + \sin x - 1) \cdot \frac{1}{2} \cot x}$$

This simplifies to:
$$\beta = e^{\lim_{x 	o 0} \sin x \cdot \frac{1}{2} \cot x}$$

Since $\cot x = \frac{\cos x}{\sin x}$, the $\sin x$ terms cancel out beautifully, leaving us with:
$$\beta = e^{\lim_{x 	o 0} \frac{1}{2} \cos x}$$

Evaluating this at $x=0$, we get **$\beta = e^{\frac{1}{2}} = \sqrt{e}$**.

### The Quadratic Bridge

Now that we have our roots, $\alpha = 1$ and $\beta = \sqrt{e}$, we can construct the quadratic equation. The standard form is $(x-\alpha)(x-\beta) = 0$, which becomes:
$$(x-1)(x-\sqrt{e}) = 0$$

Expanding this, we get:
$$x^2 - (1+\sqrt{e})x + \sqrt{e} = 0$$

But wait! The problem gives us $ax^2 + bx - \sqrt{e} = 0$. Our constant term is positive, but the given one is negative. We must multiply our equation by $-1$ to align them:
$$-x^2 + (1+\sqrt{e})x - \sqrt{e} = 0$$

Now, comparing coefficients, we find **$a = -1$** and **$b = 1+\sqrt{e}$**.

### The Final Victory

We are at the finish line. We need to calculate $12 \log_e(a+b)$.

Substituting our values:
$$a+b = -1 + 1 + \sqrt{e} = \sqrt{e}$$

Thus, we need:
$$12 \log_e(\sqrt{e})$$

Since $\sqrt{e} = e^{\frac{1}{2}}$, this becomes:
$$12 \cdot \frac{1}{2} \log_e(e)$$

Since $\log_e(e) = 1$, our final answer is **$6$**. You have navigated the limits, bridged the quadratic, and arrived at the solution. Celebrate this victory!

Enter Numerical Value:

Visualized Solution

Problem Overview

Simplifying $α$

Evaluating $α$

Identifying the Form of $β$

Applying $1^{\infty}$ Limit Formula

Evaluating $β$

Forming the Quadratic Equation

Expanding and Matching Coefficients

Finding $a$ and $b$

Final Logarithmic Calculation

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

Analyzing the Setup

Unmasking $α$

The $1^{\infty}$ Dance with $β$

The Quadratic Bridge

The Final Victory

Similar Questions

Let $α (a)$ and $β (a)$ be the roots of the equation $(3 1 + a - 1) x^{2} + (1 + a - 1) x + (6 1 + a - 1) = 0$ where $a > - 1$ . Then $lim_{a \to 0^{+}} α (a)$ and $lim_{a \to 0^{+}} β (a)$ are

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If $α, β$ are the distinct roots of $x^{2} + b x + c = 0$ , then $lim_{x \to β} \frac{e ^{2 (x^{2} + b x + c)} - 1 - 2 ( x ^{2} + b x + c )}{( x - β ) ^{2}}$ is equal to:

Enter Numerical Value:

Visualized Solution

Problem Overview

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

Identifying the Form of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β

Applying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 1∞ Limit Formula

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β

Forming the Quadratic Equation

Expanding and Matching Coefficients

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Final Logarithmic Calculation

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

Analyzing the Setup

Unmasking .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

The .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 1∞ Dance with .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β

The Quadratic Bridge

The Final Victory

Similar Questions

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