Question

, find the limit or state that it does not exist. $$ \lim _{x \rightarrow 0} \frac{x \sin 2 x}{\sin \left(x^{2}\right)} $$

Short answer

The limit is 2.

Step-by-step solution

Convert to Limit Form

First, write down the expression of the limit you want to solve: \( \lim _{x \rightarrow 0} \frac{x \sin 2 x}{\sin \left(x^{2}\right)} \). Notice that both numerator and denominator approach 0 as \( x \to 0 \), creating an indeterminate form \( \frac{0}{0} \). This signals that L'Hopital's Rule could be applicable. However, let's first try to simplify the expression using series expansion or trigonometric limits.

Apply Trigonometric Limit Rule

Recall the standard limit \( \lim_{u \to 0} \frac{\sin u}{u} = 1 \). To utilize this, rewrite the original limit as: \( \lim _{x \rightarrow 0} \frac{x \sin 2x}{x^2} \cdot \frac{x^2}{\sin (x^2)} \). This separates the sine components, so we can handle them individually.

Evaluate Limits Separately

Now, focus on evaluating the separate limits. First, consider \( \lim_{x \rightarrow 0} \frac{x}{\sin(x^2)} = \lim_{x \rightarrow 0} \frac{1}{\cos(x^2)} \cdot 1 = 1 \), because \( \cos(x^2) \to 1 \) as \( x \to 0 \). Then, evaluate \( \lim_{x \rightarrow 0} \frac{x \sin 2x}{x^2} = \lim_{x \rightarrow 0} \frac{\sin 2x}{x} = 2 \), using the substitution \( u = 2x \), this limit conforms to \( \lim_{u \rightarrow 0} \frac{\sin u}{u} = 1 \) and since \( \frac{u}{x} = 2 \), it results in \( 2 \cdot 1 = 2 \).

Combine Results

Since the limit can be expressed as a product of two limits: \( \lim_{x \rightarrow 0} \frac{x \sin 2x}{x^2} = 2 \) and \( \lim_{x \rightarrow 0} \frac{x}{\sin (x^2)} = 1 \), it follows that the original limit is \( 2 \times 1 = 2 \).

Key concepts

Trigonometric Limits

When dealing with trigonometric limits, especially in expressions that involve sine or cosine functions near the point where these functions return zero, specific strategies can be useful. For example, one critical concept is the limit \( \lim_{u \to 0} \frac{\sin u}{u} = 1 \), which plays a central role in solving various trigonometric limit problems.

This rule helps us handle cases where both the numerator and denominator approach zero, like \( \frac{\sin u}{u} \). Applying this to the original problem, recognizing that \( \lim_{x \to 0} \frac{\sin(x^2)}{x^2} = 1 \) simplifies the process tremendously. Similarly, when we set \( u = 2x \), \( \lim_{x \to 0} \frac{\sin(2x)}{x} \) can also be transformed using the same idea, providing a more straightforward way to solve the limit.

In simple terms, by expressing trigonometric functions in such a way that their arguments approach zero, we significantly simplify these limits and connect them to known standard limits.

Indeterminate Forms

Indeterminate forms arise when evaluating limits, leading to results that are not immediately clear or standard. One such type is the \( \frac{0}{0} \) form, prevalent in expressions like our example \( \lim_{x \rightarrow 0} \frac{x \sin 2x}{\sin \left(x^{2}\right)} \).

Whenever you encounter a \( \frac{0}{0} \) form, it signals that the expression isn't directly solvable without further manipulation or a deeper method, such as L'Hôpital's Rule or series expansions. The \( \frac{0}{0} \) form essentially means there's a balancing act between the numerator and denominator as they both approach zero.

Therefore, assessing the complexity by converting the problem into separate, simpler components through factorization or trigonometric rule applications is crucial to untangle these forms. This exercise demonstrates that by breaking down the original indeterminate form into manageable sections such as \( \frac{x \sin 2x}{x^2} \) and \( \frac{x^2}{\sin(x^2)} \), we can resolve each part individually before multiplying everything back together.

Series Expansion

Series expansions, such as Taylor or Maclaurin series, offer another way to handle limits, especially when functions become unwieldy or approach points leading to indeterminate forms.

Expansion transforms functions into infinite sums of polynomial terms, which approximate the function's behavior close to the expansion point, such as around zero for Maclaurin series. For trigonometric functions, this method translates sine or cosine functions into a power series, simplifying the expressions considerably.

For instance, knowing \( \sin(u) \approx u - \frac{u^3}{6} + \cdots \) as \( u \to 0 \) allows us to substitute these approximations in our original function. Although we used trigonometric limits primarily in this example, series expansions can enrich understanding and offer alternative solutions, especially if you want to see inside the function's behavior near the limit point.