Question

Evaluate the limits with either L'Hôpital's rule or previously learned methods. $$ \lim _{x \rightarrow 0} \frac{\sin x-\tan x}{x^{2}} $$

Short answer

The limit is 0.

Step-by-step solution

Verify Indeterminate Form

First, check if the limit is in an indeterminate form like \( \frac{0}{0} \). For this, evaluate the limit \( \lim _{x \to 0} \frac{\sin x - \tan x}{x^2} \). \( \sin(0) = 0 \) and \( \tan(0) = 0 \), thus \( \sin x - \tan x = 0 - 0 = 0 \) and the denominator \( x^2 \to 0 \) as \( x \to 0 \), resulting in the indeterminate form \( \frac{0}{0} \).

Apply L'Hôpital's Rule

Since the limit is in the form \( \frac{0}{0} \), we can apply L'Hôpital's Rule which states that \(\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}\), provided that the derivatives exist and \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) forms persist.Compute the derivatives:- The derivative of the numerator \( f(x) = \sin x - \tan x \) is \( f'(x) = \cos x - \sec^2 x \).- The derivative of the denominator \( g(x) = x^2 \) is \( g'(x) = 2x \).

Evaluate the New Limit

Using L'Hôpital's rule, the new limit to evaluate is \( \lim _{x \to 0} \frac{\cos x - \sec^2 x}{2x} \).Substitute \( x = 0 \) into the derivatives:- \( \cos(0) = 1 \), and \( \sec^2(0) = 1 \), so the numerator becomes \( 1 - 1 = 0 \).- The denominator \(2x \) becomes \(2 \times 0 = 0 \).This results in another \( \frac{0}{0} \) indeterminate form.

Apply L'Hôpital's Rule Again

Apply L'Hôpital's Rule again to the limit \( \lim _{x \to 0} \frac{\cos x - \sec^2 x}{2x} \).Compute the second derivatives:- The second derivative of the numerator \( f'(x) = \cos x - \sec^2 x \) is \( f''(x) = -\sin x + 2\sec^2 x\tan x \).- The derivative of the denominator \( g'(x) = 2x \) is \( g''(x) = 2 \).

Evaluate the Final Limit

Finally, evaluate the new limit \( \lim _{x \to 0} \frac{-\sin x + 2\sec^2 x\tan x}{2} \).Substitute \( x = 0 \):- \( \sin(0) = 0 \).- \( \tan(0) = 0 \) and \( \sec^2(0) = 1 \), so \( 2 \times 1 \times 0 = 0 \).Thus, the limit evaluates to \( \frac{0}{2} = 0 \).

Key concepts

L'Hôpital's Rule

L'Hôpital's Rule is a powerful method in calculus for finding limits of indeterminate forms, like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). The rule is straightforward: if a limit of a ratio of two functions results in one of these indeterminate forms, you can differentiate the numerator and the denominator separately. Then, you evaluate the limit again using these new derivatives.

• First, ensure that the original limit yields an indeterminate form.
• Differentiate the numerator and the denominator individually.
• Re-evaluate the limit using these derivatives.

Remember, you can apply L'Hôpital's Rule repeatedly if the limit still results in an indeterminate form, as shown in our exercise. The rule is particularly useful because it often simplifies the limit calculation dramatically.

Indeterminate Forms

When evaluating limits, you might encounter expressions that don't immediately lead to a clear answer. These are known as indeterminate forms. Typical indeterminate forms are:

• \( \frac{0}{0} \)
• \( \frac{\infty}{\infty} \)
• \( 0\cdot\infty \)
• \( \infty - \infty \)
• \( 1^\infty \)
• \( 0^0 \)
• \( \infty^0 \)

These forms suggest ambiguity and require additional techniques, like L'Hôpital's Rule, to resolve. For example, in our exercise, the limit \( \lim _{x \to 0} \frac{\sin x - \tan x}{x^2} \) results in the indeterminate form \( \frac{0}{0} \). Detecting this form is crucial as it indicates that more work is needed to refine the limit.

Limits

The concept of a limit is at the heart of calculus. It describes the value that a function approaches as the input approaches a certain point. Limits help in understanding the behavior of functions at points of discontinuity or where they might not be explicitly defined.
For example, the limit \( \lim _{x \to 0} \frac{\sin x - \tan x}{x^2} \) explores the behavior of the function as \( x \) approaches 0, even though the algebraic expression might be undefined at \( x = 0 \). Calculating limits can help find the direction and nature of a graph near such points. Limits are essentially the foundation tools that lead to derivatives and integrals.

Derivatives

Derivatives measure how a function changes as its input changes. It is the rate of change or the slope of the function at a given point. In the context of limits and L'Hôpital's Rule, derivatives allow us to simplify and evaluate limits that yield indeterminate forms.
By finding the derivative of the numerator and the denominator of a fraction function, L'Hôpital's Rule effectively utilizes derivatives to transform a complex limit problem into a solvable one.

• The derivative of \( \sin x - \tan x \) is \( \cos x - \sec^2 x \).
• The derivative of \( x^2 \) is \( 2x \).

These derivatives were key in our exercise to simplify and ultimately evaluate the limit. Understanding derivatives thus helps in navigating the intricacies of calculus problems like the one we solved.