Evaluating Limits Using Trigonometric Identities

This problem asks us to evaluate the limit of a function involving a square root and trigonometric terms as the variable approaches zero. The specific limit we need to find is:

\(\displaystyle\lim_{x \rightarrow 0} \frac{x}{\sqrt{1−\cos 4x}}\)

Step-by-Step Limit Evaluation

Let's evaluate the given limit. First, we can use a trigonometric identity to simplify the expression inside the square root. The identity is \(1 - \cos 2\theta = 2 \sin^2 \theta\). In our case, \(\theta = 2x\), so \(2\theta = 4x\). Applying the identity, we get:

\(1 - \cos 4x = 2 \sin^2 (2x)\)

Now substitute this into the limit expression:

\(\displaystyle\lim_{x \rightarrow 0} \frac{x}{\sqrt{2 \sin^2 (2x)}}\)

Next, simplify the square root term. Remember that \(\sqrt{a^2} = |a|\).

\(\sqrt{2 \sin^2 (2x)} = \sqrt{2} \sqrt{\sin^2 (2x)} = \sqrt{2} |\sin (2x)|\)

So the limit becomes:

\(\displaystyle\lim_{x \rightarrow 0} \frac{x}{\sqrt{2} |\sin (2x)|}\)

For the limit to exist as \(x \rightarrow 0\), the left-hand limit (\(x \rightarrow 0^-\)) and the right-hand limit (\(x \rightarrow 0^+\)) must be equal.

Evaluating the Right-Hand Limit (\(x \rightarrow 0^+\))

As \(x\) approaches \(0\) from the positive side (\(x > 0\)), \(2x\) is also positive and approaches \(0\). For small positive values, \(\sin(2x)\) is positive. Therefore, \(|\sin(2x)| = \sin(2x)\) when \(x \rightarrow 0^+\).

The limit expression becomes:

\(\displaystyle\lim_{x \rightarrow 0^+} \frac{x}{\sqrt{2} \sin (2x)}\)

We can rewrite this expression to use the standard limit \(\displaystyle\lim_{y \rightarrow 0} \frac{\sin y}{y} = 1\). Let \(y = 2x\). As \(x \rightarrow 0\), \(y \rightarrow 0\).

\(\displaystyle\lim_{x \rightarrow 0^+} \frac{x}{\sqrt{2} \sin (2x)} = \frac{1}{\sqrt{2}} \displaystyle\lim_{x \rightarrow 0^+} \frac{x}{\sin (2x)}\)

Now, manipulate the fraction \(\frac{x}{\sin(2x)}\):

\(\frac{x}{\sin (2x)} = \frac{1}{\frac{\sin (2x)}{x}} = \frac{1}{\frac{\sin (2x)}{2x} \cdot 2}\)

So the limit is:

\(\frac{1}{\sqrt{2}} \displaystyle\lim_{x \rightarrow 0^+} \frac{1}{2 \cdot \frac{\sin (2x)}{2x}}\)

Using the standard limit \(\displaystyle\lim_{y \rightarrow 0} \frac{\sin y}{y} = 1\), with \(y=2x\), we have \(\displaystyle\lim_{x \rightarrow 0^+} \frac{\sin (2x)}{2x} = 1\).

Substitute this value into the limit expression:

\(\frac{1}{\sqrt{2}} \cdot \frac{1}{2 \cdot 1} = \frac{1}{2\sqrt{2}}\)

Evaluating the Left-Hand Limit (\(x \rightarrow 0^-\))

As \(x\) approaches \(0\) from the negative side (\(x < 0\)), \(2x\) is also negative and approaches \(0\). For small negative values, \(\sin(2x)\) is negative. Therefore, \(|\sin(2x)| = -\sin(2x)\) when \(x \rightarrow 0^-\).

The limit expression becomes:

\(\displaystyle\lim_{x \rightarrow 0^-} \frac{x}{\sqrt{2} (-\sin (2x))}\)

Using the same manipulation as before:

\(\frac{1}{\sqrt{2}} \displaystyle\lim_{x \rightarrow 0^-} \frac{x}{-\sin (2x)} = \frac{1}{\sqrt{2}} \displaystyle\lim_{x \rightarrow 0^-} \frac{1}{-\frac{\sin (2x)}{x}} = \frac{1}{\sqrt{2}} \displaystyle\lim_{x \rightarrow 0^-} \frac{1}{-2 \cdot \frac{\sin (2x)}{2x}}\)

Using the standard limit \(\displaystyle\lim_{y \rightarrow 0} \frac{\sin y}{y} = 1\), with \(y=2x\), we have \(\displaystyle\lim_{x \rightarrow 0^-} \frac{\sin (2x)}{2x} = 1\).

Substitute this value into the limit expression:

\(\frac{1}{\sqrt{2}} \cdot \frac{1}{-2 \cdot 1} = -\frac{1}{2\sqrt{2}}\)

Since the left-hand limit (\(-\frac{1}{2\sqrt{2}}\)) and the right-hand limit (\(\frac{1}{2\sqrt{2}}\)) are not equal, the two-sided limit \(\displaystyle\lim_{x \rightarrow 0} \frac{x}{\sqrt{1−\cos 4x}}\) does not exist.

However, if the question is interpreted in a specific context where only the positive side is considered (e.g., in some physics problems where \(x\) represents a physical quantity always positive), or if there is another convention implied, the right-hand limit value is \(\frac{1}{2\sqrt{2}}\).

Revision Table: Key Trigonometric Identities and Limits

Additional Information on Limits and Absolute Values

When evaluating limits involving square roots of squared terms, like \(\sqrt{\sin^2(ax)}\), it is crucial to use the absolute value, as \(\sqrt{y^2} = |y|\). The absolute value function \(|y|\) is defined as:

• \(|y| = y\) if \(y \ge 0\)
• \(|y| = -y\) if \(y < 0\)

When evaluating a two-sided limit as \(x \rightarrow c\), we must check if the limit from the left (\(x \rightarrow c^-\)) equals the limit from the right (\(x \rightarrow c^+\)). If these one-sided limits are equal to a finite value \(L\), then the overall limit exists and is equal to \(L\). If the one-sided limits are not equal, or if either one-sided limit is infinite, then the overall limit does not exist.

In this problem, because \(\sin(2x)\) changes sign as \(x\) crosses 0, the term \(|\sin(2x)|\) behaves differently for \(x > 0\) and \(x < 0\), leading to different right-hand and left-hand limits.