Question

Find the value of \( \displaystyle \lim_{h \to 0} \frac{(a+h)^2 \sin(a+h) - a^2 \sin a}{h} \)

• A. \( -a^2 \sin a \)
• B. 0
• C. 1
• D. \( a^2 \cos a + 2a \sin a \)

Hint:

Limits of the form \( \frac{f(a+h)-f(a)}{h} \) directly represent the derivative \( f'(a) \). Identify the function first to simplify quickly.

Answer 1

The Correct Option is D

Step 1: Recognize the limit form.
The given limit is of the form:
\[ \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}. \]
This represents the derivative \( f'(a) \), where:
\[ f(x) = x^2 \sin x. \]

Step 2: Rewrite the expression accordingly.
Thus, the limit becomes:
\[ f'(a), \quad \text{where } f(x) = x^2 \sin x. \]

Step 3: Differentiate using product rule.
Using the product rule:
\[ \frac{d}{dx}(uv) = u'v + uv', \]
where \( u = x^2 \) and \( v = \sin x \).

Step 4: Compute derivatives of individual functions.
\[ u' = 2x, \quad v' = \cos x. \]

Step 5: Apply product rule.
\[ f'(x) = 2x \sin x + x^2 \cos x. \]

Step 6: Evaluate at \( x = a \).
\[ f'(a) = 2a \sin a + a^2 \cos a. \]

Step 7: Final simplification.
Rewriting:
\[ f'(a) = a^2 \cos a + 2a \sin a. \]
Final Answer:
\[ \boxed{a^2 \cos a + 2a \sin a}. \]