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(B) Given the limit expression: $L = \mathop {Limit}\limits_{h 	o 0} \frac{f(x + 3h) - f(x - 2h)}{h}$.Since $f(x)$ is differentiable,we can use the d

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$5f'(x)$

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(B) Given the limit expression: $L = \mathop {Limit}\limits_{h 	o 0} \frac{f(x + 3h) - f(x - 2h)}{h}$.Since $f(x)$ is differentiable,we can use the definition of the derivative or $L$'Hopital's rule.Using $L$'Hopital's rule (as it is a $0/0$ form):$L = \mathop {Limit}\limits_{h 	o 0} \frac{\frac{d}{dh}[f(x + 3h) - f(x - 2h)]}{\frac{d}{dh}[h]}$.Applying the chain rule:$L = \mathop {Limit}\limits_{h 	o 0} \frac{f'(x + 3h) \cdot 3 - f'(x - 2h) \cdot (-2)}{1}$.$L = \mathop {Limit}\limits_{h 	o 0} [3f'(x + 3h) + 2f'(x - 2h)]$.As $h 	o 0$,$f'(x + 3h) 	o f'(x)$ and $f'(x - 2h) 	o f'(x)$.Therefore,$L = 3f'(x) + 2f'(x) = 5f'(x)$.

If $f(x)$ is a differentiable function of $x$,then $\mathop {Limit}\limits_{h \to 0} \frac{f(x + 3h) - f(x - 2h)}{h} = $

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