Question

$\displaystyle\lim_{x\to 0}\frac{(2+x)\sin(2+x) - 2\sin 2}{x}$ is equal to

• A. $\sin 2$
• B. $\cos 2$
• C. 1
• D. $2\cos 2 + \sin 2$

Hint:

Recognise $\displaystyle\lim_{h\to 0}\frac{f(a+h)-f(a)}{h} = f'(a)$ to avoid direct computation of difficult limits.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Recognise the limit as the derivative of $g(t) = t\sin t$ evaluated at $t = 2$.
Step 2: Detailed Explanation:
$\displaystyle\lim_{x\to0}\frac{g(2+x)-g(2)}{x} = g'(2)$ where $g(t)=t\sin t$.
$g'(t) = \sin t + t\cos t$, so $g'(2) = \sin 2 + 2\cos 2$.
Step 3: Final Answer:
The limit equals $\sin 2 + 2\cos 2$.