Key Concepts and Formulas

• Limits at a point: Understanding how to evaluate the limit of a function as the variable approaches a specific value.
• Indeterminate Forms: Recognizing indeterminate forms like $\frac{0}{0}$ or $\infty \cdot 0$ and knowing techniques to resolve them.
• Conjugate Multiplication: A technique used to rationalize the numerator or denominator of an expression involving square roots, often used to resolve $\frac{0}{0}$ indeterminate forms.
• Trigonometric Identities: Knowledge of fundamental trigonometric identities, such as $\sin^2 x + \cos^2 x = 1$, and half-angle formulas like $1 - \cos x = 2 \sin^2 \frac{x}{2}$.
• Standard Limits: Familiarity with standard limits like $\lim_{x \to 0} \frac{\sin x}{x} = 1$ and $\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$.

Step-by-Step Solution

Let the given limit be $L$. $L = \lim _{x \rightarrow 0} \operatorname{cosec} x\left(\sqrt{2 \cos ^2 x+3 \cos x}-\sqrt{\cos ^2 x+\sin x+4}\right)$

Step 1: Rewrite the expression and identify the indeterminate form. We can rewrite $\operatorname{cosec} x$ as $\frac{1}{\sin x}$. $L = \lim _{x \rightarrow 0} \frac{\sqrt{2 \cos ^2 x+3 \cos x}-\sqrt{\cos ^2 x+\sin x+4}}{\sin x}$ As $x \rightarrow 0$, $\cos x \rightarrow 1$ and $\sin x \rightarrow 0$. The numerator approaches $\sqrt{2(1)^2 + 3(1)} - \sqrt{(1)^2 + 0 + 4} = \sqrt{2+3} - \sqrt{1+4} = \sqrt{5} - \sqrt{5} = 0$. The denominator approaches $\sin 0 = 0$. Thus, the limit is in the indeterminate form $\frac{0}{0}$.

Step 2: Multiply the numerator and denominator by the conjugate of the numerator. To resolve the $\frac{0}{0}$ indeterminate form, we multiply by the conjugate of the expression in the numerator. $L = \lim _{x \rightarrow 0} \frac{\left(\sqrt{2 \cos ^2 x+3 \cos x}-\sqrt{\cos ^2 x+\sin x+4}\right)}{\sin x} \times \frac{\left(\sqrt{2 \cos ^2 x+3 \cos x}+\sqrt{\cos ^2 x+\sin x+4}\right)}{\left(\sqrt{2 \cos ^2 x+3 \cos x}+\sqrt{\cos ^2 x+\sin x+4}\right)}$ $L = \lim _{x \rightarrow 0} \frac{(2 \cos ^2 x+3 \cos x) - (\cos ^2 x+\sin x+4)}{\sin x \left(\sqrt{2 \cos ^2 x+3 \cos x}+\sqrt{\cos ^2 x+\sin x+4}\right)}$

Step 3: Simplify the numerator. $L = \lim _{x \rightarrow 0} \frac{2 \cos ^2 x+3 \cos x - \cos ^2 x - \sin x - 4}{\sin x \left(\sqrt{2 \cos ^2 x+3 \cos x}+\sqrt{\cos ^2 x+\sin x+4}\right)}$ $L = \lim _{x \rightarrow 0} \frac{\cos ^2 x+3 \cos x - \sin x - 4}{\sin x \left(\sqrt{2 \cos ^2 x+3 \cos x}+\sqrt{\cos ^2 x+\sin x+4}\right)}$

Step 4: Rearrange the terms in the numerator to facilitate factorization. We can group terms in the numerator to try and factor them. $L = \lim _{x \rightarrow 0} \frac{(\cos ^2 x - 1) + (3 \cos x - \sin x - 3)}{\sin x \left(\sqrt{2 \cos ^2 x+3 \cos x}+\sqrt{\cos ^2 x+\sin x+4}\right)}$ This rearrangement doesn't immediately simplify well. Let's try another approach for the numerator. Consider the numerator: $\cos^2 x + 3\cos x - \sin x - 4$. We know that as $x \to 0$, $\cos x \approx 1 - \frac{x^2}{2}$ and $\sin x \approx x$. So, $\cos^2 x \approx (1 - \frac{x^2}{2})^2 \approx 1 - x^2$. The numerator is approximately $(1 - x^2) + 3(1 - \frac{x^2}{2}) - x - 4 = 1 - x^2 + 3 - \frac{3x^2}{2} - x - 4 = -\frac{5x^2}{2} - x$. This is still not straightforward.

Let's try factoring the numerator: $\cos^2 x + 3 \cos x - \sin x - 4$. We can rewrite $\cos^2 x - 4$ as $(\cos x - 2)(\cos x + 2)$. This doesn't seem helpful. Let's rewrite the numerator as: $(\cos^2 x - 1) + 3\cos x - \sin x - 3$. Using $\cos^2 x - 1 = -\sin^2 x$. Numerator: $-\sin^2 x + 3\cos x - \sin x - 3$. This still doesn't look easy to factor with $\sin x$ in the denominator.

Let's re-examine the expression after conjugate multiplication: $\frac{\cos ^2 x+3 \cos x-\sin x-4}{\sin x \left(\sqrt{2 \cos ^2 x+3 \cos x}+\sqrt{\cos ^2 x+\sin x+4}\right)}$ Let's focus on the term $\cos^2 x + 3 \cos x - 4$. We can treat this as a quadratic in $\cos x$. Let $y = \cos x$. Then $y^2 + 3y - 4 = (y+4)(y-1)$. So, $\cos^2 x + 3 \cos x - 4 = (\cos x + 4)(\cos x - 1)$. The numerator becomes $(\cos x + 4)(\cos x - 1) - \sin x$.

Step 5: Substitute the factored form and use trigonometric identities. $L = \lim _{x \rightarrow 0} \frac{(\cos x+4)(\cos x-1)-\sin x}{\sin x \left(\sqrt{2 \cos ^2 x+3 \cos x}+\sqrt{\cos ^2 x+\sin x+4}\right)}$ Now, we use the identity $\cos x - 1 = -2 \sin^2 \frac{x}{2}$ and $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$. $L = \lim _{x \rightarrow 0} \frac{(\cos x+4)(-2 \sin^2 \frac{x}{2}) - 2 \sin \frac{x}{2} \cos \frac{x}{2}}{\left(2 \sin \frac{x}{2} \cos \frac{x}{2}\right) \left(\sqrt{2 \cos ^2 x+3 \cos x}+\sqrt{\cos ^2 x+\sin x+4}\right)}$

Step 6: Factor out common terms in the numerator and simplify. We can factor out $-2 \sin \frac{x}{2}$ from the numerator. $L = \lim _{x \rightarrow 0} \frac{-2 \sin \frac{x}{2} \left[ (\cos x+4) \sin \frac{x}{2} + \cos \frac{x}{2} \right]}{2 \sin \frac{x}{2} \cos \frac{x}{2} \left(\sqrt{2 \cos ^2 x+3 \cos x}+\sqrt{\cos ^2 x+\sin x+4}\right)}$ Cancel out $2 \sin \frac{x}{2}$ from the numerator and denominator. $L = \lim _{x \rightarrow 0} \frac{-\left[ (\cos x+4) \sin \frac{x}{2} + \cos \frac{x}{2} \right]}{\cos \frac{x}{2} \left(\sqrt{2 \cos ^2 x+3 \cos x}+\sqrt{\cos ^2 x+\sin x+4}\right)}$

Step 7: Evaluate the limit by substituting $x=0$. As $x \rightarrow 0$: $\cos x \rightarrow 1$ $\sin \frac{x}{2} \rightarrow \sin 0 = 0$ $\cos \frac{x}{2} \rightarrow \cos 0 = 1$ $\sqrt{2 \cos ^2 x+3 \cos x} \rightarrow \sqrt{2(1)^2 + 3(1)} = \sqrt{5}$ $\sqrt{\cos ^2 x+\sin x+4} \rightarrow \sqrt{(1)^2 + 0 + 4} = \sqrt{5}$

Substitute these values into the expression: $L = \frac{-\left[ (1+4)(0) + 1 \right]}{1 \left(\sqrt{5}+\sqrt{5}\right)}$ $L = \frac{-[0 + 1]}{1 (2\sqrt{5})}$ $L = \frac{-1}{2\sqrt{5}}$

Common Mistakes & Tips

• Algebraic Errors: Be extremely careful with algebraic manipulations, especially when expanding squares and dealing with negative signs.
• Incorrect Application of Identities: Ensure trigonometric identities are applied correctly, particularly the half-angle formulas and the $\sin x$ double-angle formula.
• Ignoring the Denominator: When simplifying the numerator, do not forget the denominator, which is crucial for the final limit evaluation. The denominator contains the term $\left(\sqrt{2 \cos ^2 x+3 \cos x}+\sqrt{\cos ^2 x+\sin x+4}\right)$ which evaluates to $2\sqrt{5}$ and is essential.
• Approximation Errors: While Taylor series approximations can be useful for understanding limits, avoid using them directly in the final steps of a formal solution unless the problem specifically allows it or it's for verification. The exact method is preferred here.

Summary

The problem involves evaluating a limit that initially presents as an indeterminate form $\frac{0}{0}$. The strategy employed is to rewrite the expression with $\sin x$ in the denominator and then use the technique of multiplying by the conjugate of the numerator to rationalize it. After simplification, the numerator is factored using a quadratic form in $\cos x$, and trigonometric identities for $\cos x - 1$ and $\sin x$ are applied. Finally, by substituting $x=0$ into the simplified expression, the limit is evaluated.

The final answer is $\boxed{-\frac{1}{2 \sqrt{5}}}$.