If f(x) = frac{x}{1+|x|},for x in mathbb{R},t | vedclass

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(B) Given $f(x) = \frac{x}{1+|x|}$.To find $f^{\prime}(0)$,we use the definition of the derivative:$f^{\prime}(0) = \lim_{h 	o 0} \frac{f(0+h) - f(0)

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$1$

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(B) Given $f(x) = \frac{x}{1+|x|}$.To find $f^{\prime}(0)$,we use the definition of the derivative:$f^{\prime}(0) = \lim_{h 	o 0} \frac{f(0+h) - f(0)}{h}$.Since $f(0) = \frac{0}{1+|0|} = 0$,we have:$f^{\prime}(0) = \lim_{h 	o 0} \frac{\frac{h}{1+|h|} - 0}{h} = \lim_{h 	o 0} \frac{1}{1+|h|}$.Now,we evaluate the left-hand and right-hand limits:Left-hand limit $(h 	o 0^-)$: $\lim_{h 	o 0^-} \frac{1}{1-h} = 1$.Right-hand limit $(h 	o 0^+)$: $\lim_{h 	o 0^+} \frac{1}{1+h} = 1$.Since both limits are equal,$f^{\prime}(0) = 1$.

If $f(x) = \frac{x}{1+|x|}$,for $x \in \mathbb{R}$,then $f^{\prime}(0)$ is equal to :

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