Problem 50 - Thomas Calculus in SI Units

Question

Use l'Hôpital'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

The limit is \(\frac{1}{2}\).

Step 1: Verify the Indeterminate Form

First, we need to verify that the given limit results in an indeterminate form such as \(\frac{0}{0}\). Substituting \(x = 0\) into both the numerator and the denominator, we have:- Numerator: \(\sin(3 \times 0) - 3 \times 0 + 0^2 = 0\)- Denominator: \(\sin(0)\sin(2 \times 0) = 0\)Thus, the expression is of the form \(\frac{0}{0}\), allowing us to apply l'Hôpital's rule.

Step 2: Differentiate the Numerator and Denominator

Apply l'Hôpital's rule by differentiating the numerator and the denominator separately:- Derivative of the numerator: \(\frac{d}{dx} (\sin(3x) - 3x + x^2) = 3\cos(3x) - 3 + 2x\)- Derivative of the denominator: \(\frac{d}{dx} (\sin(x)\sin(2x))\).To differentiate the denominator, use the product rule: \(\frac{d}{dx}[ \sin(x) \sin(2x) ] = \sin(x)\cos(2x)\cdot 2 + \cos(x)\sin(2x) = 2\sin(x)\cos(2x) + \cos(x)\sin(2x)\).

Step 3: Evaluate the Limit Again

Now, substitute \(x = 0\) into the derivatives:- Numerator after differentiation: \(3\cos(3 \times 0) - 3 + 2 \times 0 = 3\cdot 1 - 3 = 0\)- Denominator after differentiation: \(2\sin(0)\cos(0) + \cos(0)\sin(0) = 0 + 0 = 0\).Once again, the expression is \(\frac{0}{0}\), and we apply l'Hôpital's rule a second time.

Step 4: Differentiate a Second Time

Differentiate the expressions obtained in Step 2 once more:- Second derivative of the numerator: \(\frac{d}{dx} (3\cos(3x) - 3 + 2x) = -9\sin(3x) + 2\)- Second derivative of the denominator: \(\frac{d}{dx} (2\sin(x)\cos(2x) + \cos(x)\sin(2x) )\).Apply the product rule and trigonometric derivative identities:\[\frac{d}{dx}(2\sin(x)\cos(2x)) = 2(\cos(x)\cos(2x) - \sin(x)\cdot 2\sin(2x)) = 2\cos(x)\cos(2x) - 4\sin(x)\sin(2x),\]\[\frac{d}{dx}(\cos(x)\sin(2x)) = -\sin(x)\sin(2x) + 2\cos(x)\cos(2x).\]

Step 5: Simplify and Evaluate Final Limit

Combine the second derivatives from Step 4:- Denominator after second differentiation: \( 2\cos(x)\cos(2x) - 4\sin(x)\sin(2x) - \sin(x)\sin(2x) + 2\cos(x)\cos(2x) \)- Simplified Denominator: \( 4\cos(x)\cos(2x) - 5\sin(x)\sin(2x) \).Now substitute \(x = 0\) into the second derivatives:- Numerator: \(-9\sin(3 \times 0) + 2 = 2\)- Denominator: \(4\cos(0)\cos(2 \times 0) = 4 \).The limit is \(\frac{2}{4} = \frac{1}{2}\).

Key Concepts

Calculus

Calculus is a branch of mathematics that allows us to study change. It primarily deals with derivatives and integrals, which are tools to understand how things evolve and accumulate. Calculus is split into two main disciplines: differentiation and integration.

• Differentiation - This involves finding the derivative of a function, which expresses how a function changes at a specific point. In simple terms, it tells us the slope of the function's graph.
• Integration - Integration is essentially the reverse of differentiation. It involves finding a function's integral, which represents the area under its graph. This is useful for determining total accumulation, like distance traveled over time.

Calculus is extremely powerful and versatile, utilized in many fields such as physics, engineering, economics, and medicine. It's foundational for understanding complex natural phenomena, helping scientists and engineers model systems and predict future behavior.

Indeterminate Forms

Indeterminate forms occur when evaluating limits and both the numerator and the denominator tend toward a form that doesn't have a clear value. The most common indeterminate forms are \(\frac{0}{0}\) and \(\frac{\infty}{\infty}\). These forms suggest that both the numerator and denominator move towards zero or infinity, making it unclear what the result should be.

When you encounter an indeterminate form, it’s important to remember that it does not mean the limit doesn’t exist. Instead, it indicates we need further manipulation to resolve the form. Techniques such as l'Hôpital's rule, algebraic manipulation, or trigonometric identities are often used. To properly apply l'Hôpital's rule, we must confirm the function is in an indeterminate form and then differentiate the top and bottom functions separately.

Indeterminate forms highlight the intricate behavior of functions near critical points, reminding us of the need for careful analysis in calculus.

Limits Evaluation

Limits are a fundamental concept in calculus used to understand how functions behave as the input approaches a certain value. Evaluating limits helps us investigate the behavior of functions at points where they aren't necessarily defined or when behavior changes dramatically, like near a function's asymptote or discontinuity.

There are several techniques to evaluate limits:

• Direct Substitution - Try substituting the approaching value directly into the function. If this results in a finite value, then that's the limit.
• Simplification - Factor or simplify the expression to resolve indeterminate forms like \(\frac{0}{0}\).
• L'Hôpital's Rule - If the limit yields an indeterminate form, you can use l'Hôpital's Rule, differentiating the numerator and denominator separately before re-evaluating the limit.

Understanding limits is crucial for deeper studies in calculus, as they form the basis for defining derivatives and integrals. They allow us to make sense of and work with concepts of continuity and change, which lie at calculus's core.