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Chapter 2: Problem 41

Evaluate \(\lim _{h \rightarrow 0}[f(c+h)-f(c)] / h\). $$f(x)=\sin 3 x, c=\pi / 2$$

Short Answer

Expert verified
The limit is 0.

Step by step solution

01

Substitute the function into the difference quotient

Substitute \(f(x) = \sin 3x\) and \(c = \pi / 2\) into the difference quotient \([f(c+h)-f(c)] / h\), yielding \(\frac{\sin 3(\pi / 2 + h)- \sin 3 \pi / 2}{h}\).
02

Apply trigonometric identities

Use the identity for the sin of the sum to break down the first part sin 3(\pi / 2 + h) as \(\sin (3\pi / 2 + 3h)\), yielding the new difference quotient \(\frac{\sin (3\pi / 2 + 3h) - \sin 3 \pi / 2}{h}\).
03

Simplify the expression

Simplify the expression by converting \(\sin 3\pi/2\) and \(\sin (3\pi / 2 + 3h)\) as follows: \(\frac{\sin (3\pi / 2 + 3h) - (-1)}{h}\).
04

Apply the limit

As \(h \rightarrow 0\), \(3h \rightarrow 0\) and therefore, \(\sin (3\pi / 2 + 3h)\) approaches \(\sin 3\pi/2\). This results into an indeterminate form (0/0).
05

L'Hopital's Rule

Apply L'Hopital's rule, which states that the limit of the ratio of two functions, as x approaches a particular value, is equal to the limit of the ratios of their derivatives. Thus, calculate the derivative of the numerator and denominator.
06

Apply the limit to the derivative

The derivative of the numerator \(\sin (3\pi / 2 + 3h) + 1\) is \(\cos(3\pi/2 + 3h) * 3\) and the derivative of \(h\) is 1. Therefore, substitute h = 0 and evaluate the limit, giving \(\lim_{h \rightarrow 0} 3\cos(3\pi/2 + 3h)\). This simplifies to \(3 * 0 = 0\).

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