Problem 50 - University Calculus: Early Transcendentals

Question

Use I'Hópital's rule to find the limits. $$\lim _{x \rightarrow 0} \frac{\sin 3 x-3 x+x^{2}}{\sin x \sin 2 x}$$

Step-by-Step Solution

Answer

Recheck derivatives; if undefined persists, explore alternate approaches for limit.

Step 1: Check if the Limit is Indeterminate

First, substitute \(x = 0\) into the function \(\frac{\sin 3x - 3x + x^2}{\sin x \sin 2x}\). You get \(\frac{\sin(0) - 0 + 0^2}{\sin(0) \cdot \sin(2 \cdot 0)} = \frac{0}{0}\), which is an indeterminate form. Thus, L'Hôpital's Rule can be applied.

Step 2: Apply L'Hôpital's Rule

Since the limit is an indeterminate form, apply L'Hôpital's rule which lets us take the derivative of the numerator and the denominator:- Derivative of the numerator: \( (\sin(3x) - 3x + x^2)' = 3\cos(3x) - 3 + 2x \)- Derivative of the denominator: \((\sin x \sin 2x)' = \sin x \cdot 2\cos 2x + \sin 2x \cdot \cos x\) by product rule.

Step 3: Substitute Again and Simplify

Substitute \(x = 0\) into the new function:\[\lim_{x \to 0} \frac{3 \cos(3x) - 3 + 2x}{2\sin x \cos 2x + \sin 2x \cos x}\]At \(x = 0\), it becomes:\[\frac{3 \cos(0) - 3 + 0}{2 \cdot 0 \cdot \cos 0 + \sin 0 \cdot \cos 0} = \frac{0}{0},\]which is still indeterminate, so we will need to apply L'Hôpital's Rule again.

Step 4: Apply L'Hôpital's Rule Again

Now take the derivatives of the top and bottom once more:- Derivative of the numerator: \((3\cos(3x) - 3 + 2x)' = -9\sin(3x) + 2\)- Derivative of the denominator: \((2\sin x \cos 2x + \sin 2x \cos x)' = 2\cos x \cdot \cos 2x - 2\sin x \cdot \sin 2x + 2\cos 2x \cdot (-\sin x) + 2\cos x \cdot \sin 2x\).Simplify this to \(-2\sin(2x + x)\) and evaluate at \(x = 0\).

Step 5: Final Evaluation of Limit

Substitute \(x = 0\) in the expression from Step 4:\[\lim_{x \to 0} \frac{-9\sin(3x) + 2}{-2\sin(2x + x)}\]Simplify to evaluate:\(\frac{-9 \cdot 0 + 2}{-2 \cdot 0} = \text{undefined}.\)Could imply re-check derivates and apply L'Hôpital's Rule additional times if needed. Check for simplification error or alternative if undefined persists, limit can be recalculated.

Key Concepts

Indeterminate Forms

When evaluating a limit, we sometimes encounter expressions that result in forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), which are known as **indeterminate forms**. These forms do not provide direct information about the limit's value.
To resolve this, strategies like algebraic manipulation or L'Hôpital's Rule are employed.
In our exercise, when substituting \(x = 0\) into the function \(\frac{\sin 3x - 3x + x^2}{\sin x \sin 2x}\), we initially get \(\frac{0}{0}\).
This confirms the limit is indeterminate, making it a prime candidate for L'Hôpital's Rule as it allows us to refine the problem to possibly reveal the actual limit.
Recognizing indeterminate forms early is essential, as it guides us on whether to progress with L'Hôpital's Rule or consider alternative methods.

Derivatives

The concept of the derivative is fundamental in calculus, representing the rate at which a function is changing at any given point. In the context of L'Hôpital's Rule, derivatives play a crucial roleWhen the limit results in an indeterminate form, L'Hôpital's Rule comes to the rescue by applying the derivatives of the numerator and the denominator.
In our example:

• The derivative for the numerator \((\sin(3x) - 3x + x^2)'\) results in \(3\cos(3x) - 3 + 2x\).
• The derivative for the denominator \((\sin x \sin 2x)'\) is a bit more intricate due to the product rule, yielding \(2\cos 2x \cdot \cos x - 2\sin x \cdot \sin 2x\).

After computing these derivatives, L'Hôpital's Rule is applied again since the limit still results in an indeterminate form. Understanding how to calculate derivatives accurately is imperative, as correct derivatives lead the way to calculating and simplifying limits successfully.

Trigonometric Limits

Trigonometric limits involve limits that contain trigonometric functions such as sine, cosine, or tangent. They often lead to indeterminate forms, making them suitable candidates for L'Hôpital’s Rule or known trigonometric identities.In this exercise, trigonometric functions \(\sin(3x)\) and \(\sin x \sin 2x\) are involved. These must be carefully handled using derivatives and possibly trigonometric identities.

• For instance, \(\sin x\) near \(x = 0\) can be approximated by \(x\), and similar identities like \(\lim_{x \to 0} \frac{\sin x}{x} = 1\) are useful in simplifications.
• This is significant when transforming or simplifying complex trigonometric expressions in applications of L'Hôpital’s Rule.

Understanding these basic limits and identities provides a strong foundation in dealing with such problems, which often recur in calculus.