Evaluating Limits of Trigonometric Functions

The problem asks us to find the limit of a rational expression involving the sine function as \(x\) approaches \(\frac{\pi}{6}\).

The given limit is: \[\mathop {\lim }\limits_{x \to \frac{\pi }{6}} \;\frac{{2{{\sin }^2}x\; + {\rm{\;}}\sin x\; - {\rm{\;}}1}}{{2{{\sin }^2}x\; - {\rm{\;}}3\sin x\; + {\rm{\;}}1}}\] Let's analyze the expression. It is a rational function where the variable is \(\sin x\). To simplify, let \(y = \sin x\). As \(x \to \frac{\pi}{6}\), the value of \(y = \sin x\) approaches \(\sin(\frac{\pi}{6}) = \frac{1}{2}\). So, the limit can be rewritten in terms of \(y\) as: \[\mathop {\lim }\limits_{y \to \frac{1}{2}} \;\frac{{2y^2 + y - 1}}{{2y^2 - 3y + 1}}\] Now, let's substitute \(y = \frac{1}{2}\) into the numerator and the denominator to see if we get an indeterminate form.

Checking for Indeterminate Form

• Numerator at \(y = \frac{1}{2}\): \(2\left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right) - 1 = 2\left(\frac{1}{4}\right) + \frac{1}{2} - 1 = \frac{1}{2} + \frac{1}{2} - 1 = 1 - 1 = 0\)
• Denominator at \(y = \frac{1}{2}\): \(2\left(\frac{1}{2}\right)^2 - 3\left(\frac{1}{2}\right) + 1 = 2\left(\frac{1}{4}\right) - \frac{3}{2} + 1 = \frac{1}{2} - \frac{3}{2} + 1 = -1 + 1 = 0\)

Since we get the indeterminate form \(\frac{0}{0}\), we can simplify the expression by factoring the numerator and the denominator.

Factoring the Numerator and Denominator

The numerator is a quadratic expression in \(y\): \(2y^2 + y - 1\). We can factor this expression. We look for two numbers that multiply to \(2 \times -1 = -2\) and add up to \(1\) (the coefficient of \(y\)). These numbers are \(2\) and \(-1\). We can rewrite the middle term \(y\) as \(2y - y\): \(2y^2 + y - 1 = 2y^2 + 2y - y - 1\) Group terms: \((2y^2 + 2y) + (-y - 1) = 2y(y + 1) - 1(y + 1) = (2y - 1)(y + 1)\).

The denominator is also a quadratic expression in \(y\): \(2y^2 - 3y + 1\). We look for two numbers that multiply to \(2 \times 1 = 2\) and add up to \(-3\). These numbers are \(-2\) and \(-1\). We can rewrite the middle term \(-3y\) as \(-2y - y\): \(2y^2 - 3y + 1 = 2y^2 - 2y - y + 1\) Group terms: \((2y^2 - 2y) + (-y + 1) = 2y(y - 1) - 1(y - 1) = (2y - 1)(y - 1)\).

Simplifying the Limit Expression

Now substitute the factored forms back into the limit expression: \[\mathop {\lim }\limits_{y \to \frac{1}{2}} \;\frac{{(2y - 1)(y + 1)}}{{(2y - 1)(y - 1)}}\] Since \(y \to \frac{1}{2}\), \(y\) is close to \(\frac{1}{2}\) but not equal to \(\frac{1}{2}\). Therefore, \(2y - 1\) is close to but not equal to \(0\). We can cancel the common factor \((2y - 1)\) from the numerator and the denominator: \[\mathop {\lim }\limits_{y \to \frac{1}{2}} \;\frac{{y + 1}}{{y - 1}}\] Now, we can substitute \(y = \frac{1}{2}\) directly into the simplified expression:

Evaluating the Limit

Substitute \(y = \frac{1}{2}\): \[\frac{{\frac{1}{2} + 1}}{{\frac{1}{2} - 1}} = \frac{{\frac{1}{2} + \frac{2}{2}}}{{\frac{1}{2} - \frac{2}{2}}} = \frac{{\frac{3}{2}}}{{-\frac{1}{2}}}\] To divide the fractions, multiply the numerator by the reciprocal of the denominator: \[\frac{3}{2} \times \left(-\frac{2}{1}\right) = -\frac{3 \times 2}{2 \times 1} = -3\] Thus, the value of the limit is \(-3\).

Revision Table: Limit Evaluation Steps

Review the key steps taken to solve this limit problem.

Additional Information on Limit Evaluation

When evaluating limits that result in indeterminate forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), various techniques can be used. For rational functions or functions that can be factored, cancellation of common factors is a common and effective method, as demonstrated in this problem.

Other techniques for handling indeterminate forms include:

• L'Hopital's Rule: If \(\mathop {\lim }\limits_{x \to c} \;\frac{{f(x)}}{{g(x)}}\) is of the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then \(\mathop {\lim }\limits_{x \to c} \;\frac{{f(x)}}{{g(x)}} = \mathop {\lim }\limits_{x \to c} \;\frac{{f'(x)}}{{g'(x)}}\), provided the latter limit exists. For this problem, using L'Hopital's rule on the expression in terms of \(y=\sin x\): \[\mathop {\lim }\limits_{y \to \frac{1}{2}} \;\frac{{2y^2 + y - 1}}{{2y^2 - 3y + 1}}\] The derivative of the numerator with respect to \(y\) is \(4y + 1\). The derivative of the denominator with respect to \(y\) is \(4y - 3\). Applying L'Hopital's rule: \[\mathop {\lim }\limits_{y \to \frac{1}{2}} \;\frac{{4y + 1}}{{4y - 3}} = \frac{{4(\frac{1}{2}) + 1}}{{4(\frac{1}{2}) - 3}} = \frac{{2 + 1}}{{2 - 3}} = \frac{3}{-1} = -3\] This confirms the result obtained by factorization.
• Series Expansion: For limits involving trigonometric, exponential, or logarithmic functions as \(x \to 0\), Taylor or Maclaurin series expansions can be very useful. However, this method is less common for limits approaching values other than 0, unless the expression can be easily transformed.
• Algebraic Manipulation: Techniques like multiplying by the conjugate (especially for square roots) or finding a common denominator can also help in simplifying expressions before evaluating the limit.

Understanding when to apply each technique is crucial for solving limit problems effectively. For rational functions, factorization is often the most straightforward approach when direct substitution yields \(\frac{0}{0}\).