Key Concepts and Formulas

• Limit Evaluation Techniques: When dealing with limits that result in indeterminate forms (like $\frac{0}{0}$ or $\frac{\infty}{\infty}$), we often use algebraic manipulation, such as multiplying by the conjugate, or trigonometric identities.
• Trigonometric Identities: The fundamental identity $\sin^2 x + \cos^2 x = 1$ is crucial, and its rearrangement $\cos^2 x = 1 - \sin^2 x$ will be used.
• Standard Limits: The limit $\lim_{x \to 0} \frac{\sin x}{x} = 1$ or its variations can be useful, although it might not be directly applied here. However, understanding that $\lim_{x \to \pi/2} \sin x = 1$ and $\lim_{x \to \pi/2} \cos x = 0$ is essential.

Step-by-Step Solution

Let the given limit be $L$. $L = \mathop {\lim }\limits_{x \to {\pi \over 2}} \left( {{{\tan }^2}x\left( {{{(2{{\sin }^2}x + 3\sin x + 4)}^{{1 \over 2}}} - {{({{\sin }^2}x + 6\sin x + 2)}^{{1 \over 2}}}} \right)} \right)$

Step 1: Rewrite the expression and identify the indeterminate form. We can rewrite $\tan^2 x$ as $\frac{\sin^2 x}{\cos^2 x}$. As $x \to \frac{\pi}{2}$, $\sin x \to 1$ and $\cos x \to 0$. The expression inside the parenthesis becomes: $\sqrt{2(1)^2 + 3(1) + 4} - \sqrt{(1)^2 + 6(1) + 2} = \sqrt{2+3+4} - \sqrt{1+6+2} = \sqrt{9} - \sqrt{9} = 3 - 3 = 0$. So, the limit is of the form $\infty \times 0$, which is an indeterminate form.

$L = \mathop {\lim }\limits_{x \to {\pi \over 2}} \frac{\sin^2 x}{\cos^2 x} \left( \sqrt{2{{\sin }^2}x + 3\sin x + 4} - \sqrt{{{\sin }^2}x + 6\sin x + 2} \right)$

Step 2: Rationalize the expression by multiplying by the conjugate. To handle the difference of square roots, we multiply the numerator and denominator by the conjugate of the term in the parenthesis: $\sqrt{2{{\sin }^2}x + 3\sin x + 4} + \sqrt{{{\sin }^2}x + 6\sin x + 2}$.

$L = \mathop {\lim }\limits_{x \to {\pi \over 2}} \frac{\sin^2 x}{\cos^2 x} \left( \frac{(2{{\sin }^2}x + 3\sin x + 4) - ({{\sin }^2}x + 6\sin x + 2)}{\sqrt{2{{\sin }^2}x + 3\sin x + 4} + \sqrt{{{\sin }^2}x + 6\sin x + 2}} \right)$

Step 3: Simplify the numerator after rationalization. Subtract the terms inside the parenthesis in the numerator: $(2{{\sin }^2}x + 3\sin x + 4) - ({{\sin }^2}x + 6\sin x + 2) = 2{{\sin }^2}x + 3\sin x + 4 - {{\sin }^2}x - 6\sin x - 2 = {{\sin }^2}x - 3\sin x + 2$.

$L = \mathop {\lim }\limits_{x \to {\pi \over 2}} \frac{\sin^2 x}{\cos^2 x} \left( \frac{{{\sin }^2}x - 3\sin x + 2}{\sqrt{2{{\sin }^2}x + 3\sin x + 4} + \sqrt{{{\sin }^2}x + 6\sin x + 2}} \right)$

Step 4: Factor the quadratic expression in the numerator and use $\cos^2 x = 1 - \sin^2 x$. The quadratic ${{\sin }^2}x - 3\sin x + 2$ can be factored as $({\sin x} - 1)({\sin x} - 2)$. Also, we can replace $\cos^2 x$ with $1 - \sin^2 x = (1 - \sin x)(1 + \sin x)$.

$L = \mathop {\lim }\limits_{x \to {\pi \over 2}} \frac{\sin^2 x}{(1 - \sin x)(1 + \sin x)} \left( \frac{({\sin x} - 1)({\sin x} - 2)}{\sqrt{2{{\sin }^2}x + 3\sin x + 4} + \sqrt{{{\sin }^2}x + 6\sin x + 2}} \right)$

Step 5: Cancel out common terms and simplify. Notice that $(\sin x - 1)$ in the numerator cancels with $(1 - \sin x)$ in the denominator, leaving a factor of $-1$.

$L = \mathop {\lim }\limits_{x \to {\pi \over 2}} \frac{\sin^2 x}{-(1 + \sin x)} \left( \frac{({\sin x} - 2)}{\sqrt{2{{\sin }^2}x + 3\sin x + 4} + \sqrt{{{\sin }^2}x + 6\sin x + 2}} \right)$

$L = \mathop {\lim }\limits_{x \to {\pi \over 2}} \frac{-\sin^2 x ({\sin x} - 2)}{(1 + \sin x)(\sqrt{2{{\sin }^2}x + 3\sin x + 4} + \sqrt{{{\sin }^2}x + 6\sin x + 2})}$

Step 6: Evaluate the limit by substituting $x = \frac{\pi}{2}$. As $x \to \frac{\pi}{2}$, we have $\sin x \to 1$. The numerator approaches $-(1)^2 (1 - 2) = -(1)(-1) = 1$. The denominator approaches $(1 + 1)(\sqrt{2(1)^2 + 3(1) + 4} + \sqrt{(1)^2 + 6(1) + 2})$. Denominator $\to (2)(\sqrt{2+3+4} + \sqrt{1+6+2}) = (2)(\sqrt{9} + \sqrt{9}) = (2)(3 + 3) = (2)(6) = 12$.

$L = \frac{1}{12}$

Common Mistakes & Tips

• Indeterminate Form Recognition: Always check the form of the limit first. If it's indeterminate (like $\frac{0}{0}$ or $\infty \times 0$), proceed with simplification techniques.
• Algebraic Errors: Be meticulous with algebraic manipulations, especially when rationalizing or factoring. A single sign error can lead to the wrong answer.
• Trigonometric Substitution: Remember the identity $\cos^2 x = 1 - \sin^2 x$ and how to factor it as $(1 - \sin x)(1 + \sin x)$. This is crucial for cancelling terms.

Summary

The given limit was an indeterminate form of type $\infty \times 0$. We first rewrote $\tan^2 x$ as $\frac{\sin^2 x}{\cos^2 x}$ and then rationalized the expression by multiplying by the conjugate of the difference of square roots. After simplifying the numerator and factoring the quadratic term, we used the identity $\cos^2 x = 1 - \sin^2 x$ to introduce terms that could be cancelled. Finally, by substituting $x = \frac{\pi}{2}$ (which implies $\sin x = 1$), we evaluated the limit.

The final answer is $\boxed{\frac{1}{12}}$.