Problem 47 - Calculus - AP Edition

Question

(a) Show that \(\lim _{h \rightarrow 0} \frac{\tan h}{h}=1\) (b) Use the result in part (a) to help derive the formula for the derivative of tan \(x\) directly from the definition of a derivative.

Step-by-Step Solution

Answer

(a) \( \lim_{h \to 0} \frac{\tan h}{h} = 1 \); (b) Derivative of \( \tan x \) is \( \sec^2 x \).

Step 1: Recall the Squeeze Theorem

The Squeeze Theorem is used to find the limit of a function trapped between two other functions whose limits are known and equal. It states that if \( f(h) \leq g(h) \leq f(h) \) and \( \lim_{h \to 0} f(h) = \lim_{h \to 0} f(h) = L \), then \( \lim_{h \to 0} g(h) = L \).

Step 2: Apply Angle Approximations

We know that for small angles, \( \tan(h) \approx h \). Therefore, we can apply the squeeze theorem using the known limits: \( \frac{\sin h}{h} \to 1 \) and \( \cos h \to 1 \).

Step 3: Bound \(\tan(h)\) Using Trigonometric Identities

Using the definition \(\tan(h) = \frac{\sin(h)}{\cos(h)}\), we observe that near \(h = 0\), \(\sin(h)\) and \(h\) converge similarly. Thus, \(\lim_{h \to 0} \frac{\sin h}{h} = 1\) and \(\cos(h)\) approaches 1 gives us \( \lim_{h \to 0} \frac{\tan h}{h} = 1 \).

Step 4: Use Result from Part (a) for Derivation

If we have \( \frac{\tan h}{h} \to 1 \), it implies the derivative of \( \tan(x) \) can be found using the definition \( \lim_{h\to 0} \frac{\tan(x+h) - \tan(x)}{h} \).

Step 5: Apply the Tan Angle Addition Formula

Utilize the identity \( \tan(x+h) = \frac{\tan(x) + \tan(h)}{1-\tan(x)\tan(h)} \) to substitute into the derivative definition and simplify.

Step 6: Simplify and Take the Limit

Simplify the expression from the angle addition formula for \( \tan(x) \), so that \( \lim_{h \to 0} \) results in \(1 + \tan^2(x)\), which is \(\sec^2(x)\).

Step 7: Conclude with the Derivative of \(\tan(x)\)

Since the limit of the derivative expression simplifies to \( \sec^2(x) \), the derivative of \( \tan(x) \) is \( \sec^2(x) \).

Key Concepts

Squeeze Theorem

The Squeeze Theorem is a crucial tool in calculus for finding the limits of certain functions. When a function is difficult to evaluate directly, finding two simpler functions that trap (or "squeeze") the original function can be helpful. In mathematical terms, if we have functions \( f(h) \), \( g(h) \), and \( f(h) \) such that \( f(h) \leq g(h) \leq f(h) \) and if both \( f(h) \) functions share the same limit \( L \) as \( h \) approaches a specific value, then \( g(h) \) must also converge to \( L \) at that limit.

• The theorem is particularly useful for handling limits involving trigonometric functions at small angles.
• This approach is beneficial when direct substitution is complicated or leads to indeterminate forms.

In our problem, understanding that \( \tan(h) \approx h \) for small \( h \) allows us to use the Squeeze Theorem by recognizing familiar trigonometric limits. We use the known limit \( \lim_{h \to 0} \frac{\sin h}{h} = 1 \) and \( \cos h \to 1 \). By bounding \( \tan(h) \) between correct functions using these known results, we employ the Squeeze Theorem to prove \( \lim_{h \to 0} \frac{\tan h}{h} = 1 \).

trigonometric limits

Trigonometric limits serve as foundational tools, especially when working with limits as angles approach zero. Some of the most well-known limits involve the functions \( \sin \), \( \cos \), and \( \tan \).
For instance, the limit \( \lim_{h \to 0} \frac{\sin h}{h} = 1 \) is widely useful because as \( h \) gets very small, \( \sin(h) \) closely approximates \( h \). Similarly, \( \lim_{h \to 0} \cos h = 1 \) is naturally intuitive since the cosine of a very small angle is nearly one.

• These limits make analyzing behavior of trigonometric functions at near-zero angles simpler.
• They often help refine problems that might initially seem computationally complex into manageable steps.

Working through our exercise, these trigonometric concepts were crucial. They were applied to determine that \( \lim_{h \to 0} \frac{\tan h}{h} = 1 \), effectively using the trigonometric identity \( \tan(h) = \frac{\sin h}{\cos h} \). By using both \( \sin \) and \( \cos \) limits, we gained a significant insight into how \( \tan(h) \) behaves close to zero.

derivative of tangent

The derivative of \( \tan(x) \) gives us insights into how the tangent function changes over its domain. Deriving it directly involves using the definition of a derivative and some trigonometric identities.
The derivative is formally given by the limit expression:\[\lim_{h\to 0} \frac{\tan(x+h) - \tan(x)}{h}\]Using the tan angle addition formula, the expression \( \tan(x+h) = \frac{\tan(x) + \tan(h)}{1-\tan(x)\tan(h)} \) can be substituted into the limit definition and simplified.

• This simplification uses the earlier limit result \( \frac{\tan(h)}{h} \rightarrow 1 \).
• The process ultimately simplifies the terms to yield \( \,1 + \tan^2(x) \,\).

This expression \( 1 + \tan^2(x) \) is recognized as \( \sec^2(x) \), based on trigonometric identities. Thus, the derivative of tangent is \( \sec^2(x) \). This means that the rate of change of \( \tan(x) \) at any point \( x \) is directly related to the secant squared of that angle.