Problem 13 - Single Variable Calculus

Question

evaluate the limit using l'Hôpital's Rule if appropriate. $$ \lim _{x \rightarrow 0} \frac{\sin x-x \cos x}{\tan ^{3} x} $$

Step-by-Step Solution

Answer

The limit of the given expression as \(x \rightarrow 0\) is infinity.

Step 1: Check if the limit is in 0/0 indeterminate form

Let's first find the limit of the numerator and the denominator as \(x \rightarrow 0\). Numerator: \(\lim _{x \rightarrow 0} (\sin x-x \cos x)\) Denominator: \(\lim _{x \rightarrow 0} (\tan^3 x)\) As \(x \rightarrow 0\), \(\sin x \rightarrow 0\), \(x \cos x \rightarrow 0\), and \(\tan x \rightarrow 0\), so we have: Numerator: \(0-0=0\) Denominator: \(0^3 = 0\) This means that the limit, \(\lim _{x \rightarrow 0} \frac{\sin x - x\cos x}{\tan^3 x}\), is in the 0/0 indeterminate form. Therefore, we can proceed to apply L'Hôpital's Rule.

Step 2: Apply L'Hôpital's Rule and differentiate the numerator and the denominator

According to L'Hôpital's Rule, if \(\lim _{x \rightarrow a} \frac{f(x)}{g(x)}\) has the indeterminate form 0/0 or ±∞/±∞, then: \(\lim _{x \rightarrow a} \frac{f(x)}{g(x)} = \lim _{x \rightarrow a} \frac{f'(x)}{g'(x)}\), provided the right side limit exists. So for our limit, we need to find the derivative of the numerator and the denominator with respect to x: Numerator derivative: \( (\sin x - x\cos x)'\) Denominator derivative: \((\tan^3 x)'\) \( (\sin x - x\cos x)' = (\sin x)' - (x\cos x)'\) \( (\sin x)' = \cos x \) \( (x\cos x)' = (\cos x + x(-\sin x)) = \cos x - x\sin x\) Using the chain rule, we differentiate the denominator: \( (\tan^3 x)' = 3(\tan^2 x)(\tan x)'\) \( (\tan x)' = \frac{1}{\cos^2 x} = \sec^2 x\) So, \( (\tan^3 x)' = 3(\tan^2 x)(\sec^2 x)\) Now, applying L'Hôpital's Rule, we get a new limit expression: \(\lim_{x \rightarrow 0} \frac{\cos x - x\sin x}{3\tan^2 x\sec^2 x}\)

Step 3: Check if the new limit expression can be evaluated directly

Let's check if we can evaluate the new limit expression directly as \(x \rightarrow 0\): Numerator: \(\lim _{x \rightarrow 0} (\cos x - x\sin x) = 1 - 0 = 1\) Denominator: \(\lim _{x \rightarrow 0} (3\tan^2 x\sec^2 x) = 3(0)^2(1)^2 = 0\) The new expression now has an indeterminate form of \(1/0\). We can deduce the limit as follows: 1. As \(x \rightarrow 0\), \(\tan x \rightarrow 0\) and \(\sec x \rightarrow 1\) 2. This means that \(\tan^2 x \rightarrow 0\), since \((0)^2 = 0\) 3. However, \(\cos x - x\sin x \rightarrow 1\), which is a nonzero value. 4. As a result, when we divide \(1\) (the numerator) by \(0\) (the denominator), the expression goes to infinity.

Step 4: Evaluate the final limit

From our analysis in the previous step, the limit can be deduced as: \(\lim _{x \rightarrow 0} \frac{\sin x - x\cos x}{\tan^3 x} = \lim_{x \rightarrow 0} \frac{\cos x - x\sin x}{3\tan^2 x\sec^2 x} = \infty\) Therefore, the limit of the given expression as \(x \rightarrow 0\) is infinity.

Key Concepts

Indeterminate Forms

When evaluating limits, an indeterminate form arises when both the numerator and the denominator approach zero or both approach infinity. This makes the limit seem undefined because we're essentially trying to make sense of expressions like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). In the context of this problem, as \( x \rightarrow 0 \), the original expression \( \frac{\sin x - x \cos x}{\tan^3 x} \) presents such an indeterminate form: \( \frac{0}{0} \). Recognizing this form is crucial as it allows us to apply techniques like L'Hôpital's Rule to find the limit. It is important to note that not all \( \frac{0}{0} \) forms are indeterminate—they need to be evaluated in context, often requiring algebraic simplification or advanced calculus techniques like L'Hôpital's Rule.

Differentiation

Differentiation is a fundamental tool in calculus used to find the rate of change of a function. In this problem, we used differentiation to apply L'Hôpital's Rule, which states that if the limit \( \lim_{x \rightarrow a} \frac{f(x)}{g(x)} \) results in an indeterminate form, the limit of the derivatives \( \lim_{x \rightarrow a} \frac{f'(x)}{g'(x)} \) can be evaluated instead.

• First, for the numerator \( f(x) = \sin x - x \cos x \), we found \( f'(x) \) to be \( \cos x - x \sin x \) using basic derivative rules: the derivative of \( \sin x \) is \( \cos x \), and the product rule was applied to \( x\cos x \).
• For the denominator \( g(x) = \tan^3 x \), the derivative \( g'(x) \) was determined by using the chain rule, resulting in \( 3\tan^2 x \sec^2 x \).

By differentiating both parts, we translated the original limit problem into a more manageable form with continuous functions.

Trigonometric Limits

Trigonometric limits are a special category of problems where evaluating the limit involves trigonometric functions such as \( \sin x \), \( \cos x \), and \( \tan x \). These limits often require the use of identities and approximations, especially when approaching values where these functions take on values that lead to indeterminate forms.

• Common trigonometric limits, like \( \lim_{x \rightarrow 0} \frac{\sin x}{x} = 1 \) and \( \lim_{x \rightarrow 0} \frac{1 - \cos x}{x} = 0 \), are fundamental in evaluating more complex expressions.
• In the context of this problem, as \( x \to 0 \), both \( \sin x \) and \( \tan x \) approach 0, leading to our initial indeterminate form.
• The new expression after applying L'Hôpital's rule results in a simple evaluation of limits like \( \lim_{x \rightarrow 0} (\cos x - x\sin x) = 1 \), simplifying the resolution of the limit problem.

Understanding these basic limits and how they behave near specific points helps significantly in manipulating and solving limit problems involving trigonometric functions.