Evaluating the Trigonometric Limit

The problem asks us to find the value of the limit:

This limit involves trigonometric functions and has the form of a derivative. We can solve this problem using two main methods: by applying trigonometric identities and standard limits, or by recognizing the definition of a derivative.

Method 1: Using Trigonometric Identities and Standard Limits

We can use the trigonometric identity for the difference of squares of sines: \(\sin^2 A - \sin^2 B = \sin(A-B)\sin(A+B)\). In this expression, we have \(A = x+h\) and \(B = x\).

Applying the identity to the numerator:

\(\sin^2(x+h) - \sin^2 x = \sin((x+h)-x) \sin((x+h)+x)\)

\(= \sin(h) \sin(2x+h)\)

Now, substitute this back into the limit expression:

We can rewrite this limit as a product of two limits:

We know the standard limit \(\displaystyle\lim_{h \rightarrow 0} \frac{\sin h}{h} = 1\).

For the second limit, as \(h \rightarrow 0\), the expression \(\sin(2x+h)\) approaches \(\sin(2x+0) = \sin(2x)\) because the sine function is continuous.

Therefore, the value of the limit is:

\(= 1 \cdot \sin(2x)\)

\(= \sin(2x)\)

Method 2: Recognizing the Definition of a Derivative

The definition of the derivative of a function \(f(x)\) with respect to \(x\) is given by:

Comparing this definition with the given limit expression:

We can see that the function \(f(x)\) in this case is \(f(x) = \sin^2 x\). The limit is asking for the derivative of \(f(x) = \sin^2 x\) with respect to \(x\).

Let's find the derivative of \(f(x) = \sin^2 x\). We can rewrite \(f(x)\) as \((\sin x)^2\). Using the chain rule, which states that \(\frac{d}{dx} [u(x)]^n = n[u(x)]^{n-1} \cdot u'(x)\), with \(u(x) = \sin x\) and \(n=2\):

\(\frac{d}{dx}(\sin^2 x) = \frac{d}{dx}(\sin x)^2\)

\(= 2 (\sin x)^{2-1} \cdot \frac{d}{dx}(\sin x)\)

\(= 2 \sin x \cdot \cos x\)

Now, we use the double angle identity for sine: \(\sin(2x) = 2 \sin x \cos x\).

So, the derivative of \(\sin^2 x\) is:

\(= \sin(2x)\)

Since the given limit is the definition of the derivative of \(\sin^2 x\), the value of the limit is equal to the derivative, which is \(\sin(2x)\).

Conclusion

Both methods yield the same result. The limit \(\displaystyle\lim_{h \rightarrow 0} \frac{\sin^2(x+h)−\sin^2x}{h}\) is equal to \(\sin(2x)\).

Therefore, the correct answer is \(\sin 2x\).

Revision Table: Key Concepts for Trigonometric Limits

Additional Information on Evaluating Limits

Evaluating limits is a fundamental concept in calculus. When faced with a limit problem, especially one that results in an indeterminate form like \(\frac{0}{0}\) (which this one does if you substitute \(h=0\) directly), common strategies include:

• Algebraic Simplification: Factoring, rationalizing, or combining fractions to cancel out terms that cause the indeterminate form.
• Using Standard Limits: Recognizing patterns that match known limits (like \(\lim_{x \to 0} \frac{\sin x}{x}\), \(\lim_{x \to 0} \frac{1-\cos x}{x^2}\), etc.).
• Trigonometric Identities: Applying identities to transform the expression into a form that can be evaluated.
• L'Hôpital's Rule: If the limit is of the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), you can take the derivative of the numerator and denominator separately and evaluate the limit of the new expression. (For this problem, applying L'Hôpital's rule would also lead to the correct answer).
• Recognizing Derivative or Integral Definitions: As shown in Method 2, sometimes the limit expression is precisely the definition of a derivative or a definite integral, allowing you to use differentiation or integration techniques.

Choosing the best method often depends on the structure of the expression. For trigonometric limits, using identities and standard limits is very common. Recognizing the derivative definition provides an elegant alternative when applicable.