Question

\(\lim _{\mathrm{x} \rightarrow 0}\left(3 / \mathrm{x}^{3}\right) \sin \left(\pi^{2}+2 \mathrm{x}\right)-\left(3 / \mathrm{x}^{3}\right) \sin \left(\pi^{2}+\mathrm{x}\right)\) \(-\left(1 / \mathrm{x}^{3}\right) \sin \left(\pi^{2}+3 \mathrm{x}\right)+\left(1 / \mathrm{x}^{3}\right) \sin [\pi(1+\pi)]=?\) (a) \(\operatorname{Cos} \pi^{2}\) (b) \(-\operatorname{Cos} \pi^{2}\) (c) \(-\pi\) (d) \(\pi\)

Short answer

The problem does not provide enough information to determine the exact limit, but using the given choices, the answer is either (C) -π or (D) π.

Step-by-step solution

Factor out terms from the expression

Given the expression, first find the common terms and factor them out. This will simplify the expression and make it easier to work with: \[ \lim _{x \rightarrow 0}\left[ (\frac{3}{x^{3}}) (\sin(\pi^{2}+2x)-\sin(\pi^{2}+x))-(\frac{1}{x^{3}}) (\sin(\pi^{2}+3x)-\sin(\pi(1+\pi))) \right] \]

Use a trigonometric identity to simplify the expression

Since the expression consists of sin(A) - sin(B) functions, we can use the sum-to-product identity to simplify it: \[ \sin(A) - \sin(B) = 2 \cos(\frac{A+B}{2}) \sin(\frac{A-B}{2}) \] Apply this identity to the given expression: \[ \lim _{x \rightarrow 0}\left[ (\frac{3}{x^{3}}) (2 \cos(\frac{\pi^{2}+2x+\pi^{2}+x}{2}) \sin(\frac{\pi^{2}+2x-\pi^{2}-x}{2})) - (\frac{1}{x^{3}}) (2\cos(\frac{\pi^{2}+3x+\pi(1+\pi)}{2}) \sin(\frac{\pi^{2}+3x-\pi(1+\pi)}{2})) \right] \]

Simplify the expression

Simplify the expression within the limits: \[ \lim _{x \rightarrow 0}\left[ (\frac{6}{x^{3}})\cos \left(\frac{4\pi^{2}+4x}{2}\right)\sin \left(\frac{2x-x}{2}\right) - (\frac{2}{x^{3}})\cos \left(\frac{2\pi^{2}+3x+\pi^{2}+\pi^{3}}{2}\right)\sin \left(\frac{2x-2\pi^{2}}{2}\right) \right] \]

Evaluate the limits

Now, we can take the limit as x approaches 0: \[ \frac{6}{0} \cos(\pi^{2}) \cdot 0 - \frac{2}{0} \cos(\frac{\pi^{3}}{2}) \cdot \sin(-\pi^{2}) \] As you can see, the limit does not exist as there is a division by 0. However, we can analyze the options given in the problem to eliminate impossible solutions. Since the denominator contains x^3, the limit cannot be a constant or a cosine function. The only options left are: (C) -π and (D) π. Though the exact limit cannot be found using basic limit rules, with the given choices, the answer is either (C) or (D). However, the problem does not provide enough information for determining which one is correct in this case.

Key concepts

Limits Approaching Zero

When we talk about limits approaching zero, we are considering how a function behaves as the input value gets infinitely close to zero. It's a fundamental concept in calculus, especially for understanding the behavior of functions near a specific point. For trigonometric functions, certain limits are essential to know: for example, \( \lim_{x \to 0} \frac{\sin(x)}{x} = 1 \) and \( \lim_{x \to 0} \frac{\cos(x) - 1}{x} = 0 \).

These limits are especially important because they often arise in problems involving indeterminate forms, like 0/0. In calculus, evaluating such limits requires using special techniques or identities to rewrite the function so that the limit can be computed in a straightforward manner. Often, this involves simplifying expressions or applying known limits of sine and cosine as x approaches zero.

Sum-to-Product Trigonometric Identities

Sum-to-product identities are trigonometric formulas that convert the sum or difference of two sine or cosine functions into a product of trigonometric functions. They are particularly useful when dealing with expressions involving trigonometric functions that need to be simplified. One such identity that often comes into play is \( \sin(A) - \sin(B) = 2 \cos\left(\frac{A + B}{2}\right) \sin\left(\frac{A - B}{2}\right) \).

By applying these identities, we can turn troublesome subtraction expressions into more manageable products, which can make finding limits much easier. In the case of the exercise given, the sum-to-product identity is used to transform the difference of sines into a product involving cosine and sine, paving the way to simplify the limit and possibly evaluate it.

Sine and Cosine Limits

Sine and cosine limits are a staple in calculus, particularly when dealing with periodic functions as they oscillate between their maximum and minimum values. Knowing the limits of sine and cosine as x approaches zero is crucial because these values often serve as building blocks in more complex limit problems.

Two key limits are \( \lim_{x \to 0} \sin(x) = 0 \) and \( \lim_{x \to 0} \cos(x) = 1 \), which hold true for any approach to zero, whether from the positive or negative side. In exercises where sine and cosine functions are involved in limits approaching zero, these identities allow us to replace the trigonometric functions with their limiting values to evaluate the overall limit.

Evaluating Limits

Evaluating limits is the process of finding the value that a function approaches as the input approaches a certain point. This process is fundamental to calculus and has applications ranging from determining the continuity of a function to calculating derivatives and integrals.

When evaluating limits, one may encounter different types of indeterminate forms such as 0/0, ∞/∞, or 0 * ∞. To resolve these, techniques like factoring, rationalizing, expanding, or applying L'Hôpital's Rule may be used. Additionally, recognizing patterns and applying known limits of basic functions can greatly simplify the evaluation process. In problems where the limit cannot be found directly or does not exist, understanding the behavior of the function can still provide valuable insight, as illustrated in the given exercise.