lim x→0 (x + 2sin x) / (√(x² + 2sin x + 1) - √(sin² x - x + 1)) is : (A) 3 (B) 2 (C) 6 (D) 1

Q: Why can't we just use L'Hospital's Rule immediately? It seems faster.

While L'Hospital's Rule is a valid path, differentiating $\sqrt{x^2 + 2\sin x + 1}$ involves the chain rule, leading to complex expressions that are prone to algebraic errors. Rationalization keeps the expression cleaner and builds your intuition for how functions behave near zero.

Q: Wait, why does $\frac{\sin^2 x}{x}$ go to 0 as $x \to 0$?

Think of it as $\frac{\sin x}{x} \cdot \sin x$. We know $\frac{\sin x}{x}$ approaches 1, and $\sin x$ approaches 0. Since $1 \cdot 0 = 0$, the entire term vanishes. It is a beautiful example of how different parts of a function contribute differently to the limit.

Q: Do I always have to divide by $x$ after rationalizing?

Not always, but it is a standard strategy when you have a $\frac{0}{0}$ form. Dividing by $x$ is essentially 'extracting' the factor that is causing the zero in both the numerator and denominator, allowing you to evaluate the remaining parts safely.

Q: What if the conjugate makes the numerator even more complicated?

It might look intimidating, but remember that the conjugate is just a multiplier of 1. You aren't changing the value of the expression; you are just changing its 'outfit' to reveal the hidden limit. Trust the algebra!

Q: Is there a way to use Taylor Series expansion here?

Absolutely! You could expand $\sin x \approx x - \frac{x^3}{6}$ and the square roots using the binomial expansion $(1+u)^{1/2} \approx 1 + \frac{u}{2}$. It is a very elegant, high-level approach, but rationalization is often safer for JEE exams to avoid missing higher-order terms.

Question

$\lim_{x 	o 0} \frac{x + 2\sin x}{\sqrt{x^2 + 2\sin x + 1} - \sqrt{\sin^2 x - x + 1}}$ is :

Options:

(A) 3

(B) 2

(C) 6

(D) 1

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are tasked with evaluating the limit:
$$\lim_{x 	o 0} \frac{x + 2\sin x}{\sqrt{x^2 + 2\sin x + 1} - \sqrt{\sin^2 x - x + 1}}$$

First, we check the behavior of the expression at $x = 0$. Substituting the value, we obtain:
$$\frac{0 + 2\sin(0)}{\sqrt{0+0+1} - \sqrt{0-0+1}} = \frac{0}{1 - 1} = \frac{0}{0}$$

This confirms we are dealing with an **indeterminate form**. We must manipulate the expression to resolve this ambiguity.

### The Power of the Conjugate

When square roots appear in the denominator, the most effective strategy is to rationalize the expression. We multiply the numerator and the denominator by the conjugate:
$$\sqrt{x^2 + 2\sin x + 1} + \sqrt{\sin^2 x - x + 1}$$

Applying the identity $(a-b)(a+b) = a^2 - b^2$, the denominator simplifies significantly:
$$(x^2 + 2\sin x + 1) - (\sin^2 x - x + 1) = x^2 - \sin^2 x + 2\sin x + x$$

The constants $+1$ and $-1$ cancel out, leaving us with a much cleaner expression.

### Simplifying the Landscape

Our expression now takes the form:
$$\lim_{x 	o 0} \frac{(x + 2\sin x)(\sqrt{x^2 + 2\sin x + 1} + \sqrt{\sin^2 x - x + 1})}{x^2 - \sin^2 x + 2\sin x + x}$$

To resolve the limit, we divide both the numerator and the denominator by $x$. We utilize the standard limit $\lim_{x 	o 0} \frac{\sin x}{x} = 1$.

For the numerator, dividing by $x$ yields:
$$\left(1 + 2\frac{\sin x}{x}ight) \cdot \left(\sqrt{x^2 + 2\sin x + 1} + \sqrt{\sin^2 x - x + 1}ight)$$

As $x 	o 0$, this expression approaches:
$$(1 + 2(1)) \cdot (\sqrt{1} + \sqrt{1}) = 3 \cdot 2 = 6$$

### The Final Reveal

Now, we examine the denominator after dividing by $x$:
$$\frac{x^2 - \sin^2 x + 2\sin x + x}{x} = x - \frac{\sin^2 x}{x} + 2\frac{\sin x}{x} + 1$$

As $x 	o 0$, the terms behave as follows:
$$0 - 0 + 2(1) + 1 = 3$$

Finally, we divide the limit of the numerator by the limit of the denominator:
$$\frac{6}{3} = 2$$

The value of the limit is **2**.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: $lim_{x \to 0} \frac{x + 2 s i n x}{x ^{2} + 2 s i n x + 1 - s i n ^{2} x - x + 1}$ is :

Select Answer:

Visualized Solution

Checking the Indeterminate Form

Strategy: Rationalization

Multiplying by the Conjugate

Expanding the Denominator

Simplifying the Denominator

Strategy: Standard Limits

Executing the Division

Applying the Limit

Final Calculation

Conclusion & Key Takeaway

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

Analyzing the Setup

The Power of the Conjugate

Simplifying the Landscape

The Final Reveal

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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; } limx→0​x2+2sinx+1​−sin2x−x+1​x+2sinx​ is :