My attempt of this questions ,actually the ans only a but i am getting a,b,c how?please help
$$\begin{array}{rcl} f'(k) &=& \displaystyle \lim_{h\to0} \frac{f(k+h)-f(k)}{h} \\ &=& \displaystyle \lim_{h\to0} \frac{f(k)+f(h)-f(k)}{h} \\ &=& \displaystyle \lim_{h\to0} \frac{f(h)}{h} \\ &=& \displaystyle \lim_{h\to0} \frac{\sin(h)g(h)}{h} \\ &=& \displaystyle \left( \lim_{h\to0} \frac{\sin(h)}{h} \right) \left( \lim_{h\to0} g(h) \right) \\ &=& 1 \times k \\ &=& k \end{array}$$
Note that $\displaystyle \lim_{h\to0} g(h)$ exists and is equal to $g(0)$ because $g$ is continuous.
(b) and (c) are wrong for a simple reason: $f'(k)$ is a constant and does not depend on $x$.
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$$\begin{array}{rcl} f'(k) &=& \displaystyle \lim_{h\to0} \frac{f(k+h)-f(k)}{h} \\ &=& \displaystyle \lim_{h\to0} \frac{f(k)+f(h)-f(k)}{h} \\ &=& \displaystyle \lim_{h\to0} \frac{f(h)}{h} \\ &=& \displaystyle \lim_{h\to0} \frac{\sin(h)g(h)}{h} \\ &=& \displaystyle \left( \lim_{h\to0} \frac{\sin(h)}{h} \right) \left( \lim_{h\to0} g(h) \right) \\ &=& 1 \times k \\ &=& k \end{array}$$
Note that $\displaystyle \lim_{h\to0} g(h)$ exists and is equal to $g(0)$ because $g$ is continuous.
(b) and (c) are wrong for a simple reason: $f'(k)$ is a constant and does not depend on $x$.