If f(x) = frac{sin(frac{pi x}{4})}{x + 1},the | vedclass

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(D) Let $L = \lim_{h 	o 0} \frac{f(1 + h) - f(1)}{h^2 + 2h}$.Using the definition of the derivative,$\lim_{h 	o 0} \frac{f(1+h) - f(1)}{h} = f'(1)$.

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$\frac{\pi - 4}{16\sqrt{2}}$

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(D) Let $L = \lim_{h 	o 0} \frac{f(1 + h) - f(1)}{h^2 + 2h}$.Using the definition of the derivative,$\lim_{h 	o 0} \frac{f(1+h) - f(1)}{h} = f'(1)$.We can rewrite the limit as $\lim_{h 	o 0} \left( \frac{f(1+h) - f(1)}{h} \cdot \frac{1}{h+2} ight) = f'(1) \cdot \frac{1}{2} = \frac{f'(1)}{2}$.Given $f(x) = \frac{\sin(\frac{\pi x}{4})}{x + 1}$,we use the quotient rule: $f'(x) = \frac{(x+1) \cdot \cos(\frac{\pi x}{4}) \cdot \frac{\pi}{4} - \sin(\frac{\pi x}{4}) \cdot 1}{(x+1)^2}$.At $x = 1$,$f'(1) = \frac{(1+1) \cdot \cos(\frac{\pi}{4}) \cdot \frac{\pi}{4} - \sin(\frac{\pi}{4})}{(1+1)^2} = \frac{2 \cdot \frac{1}{\sqrt{2}} \cdot \frac{\pi}{4} - \frac{1}{\sqrt{2}}}{4} = \frac{\frac{\pi}{2\sqrt{2}} - \frac{1}{\sqrt{2}}}{4} = \frac{\pi - 2}{8\sqrt{2}}$.Thus,the limit is $\frac{f'(1)}{2} = \frac{\pi - 2}{16\sqrt{2}}$.

If $f(x) = \frac{\sin(\frac{\pi x}{4})}{x + 1}$,then $\lim_{h \to 0} \frac{f(1 + h) - f(1)}{h^2 + 2h}$ is -

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