If f(x) = begin{cases} 3 + x; & x geqslant 0 | vedclass

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(D) To find $\lim_{x 	o 0} f(f(x))$,we evaluate the left-hand limit $(LHL)$ and right-hand limit $(RHL)$ at $x = 0$.For the $LHL$ $(x 	o 0^-)$: As $

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does not exist

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(D) To find $\lim_{x 	o 0} f(f(x))$,we evaluate the left-hand limit $(LHL)$ and right-hand limit $(RHL)$ at $x = 0$.For the $LHL$ $(x 	o 0^-)$: As $x 	o 0^-$,$f(x) = 2 - 3x$. Since $x$ is slightly less than $0$,$f(x)$ is slightly greater than $2$. Let $u = f(x)$. As $x 	o 0^-$,$u 	o 2^+$. Since $u > 0$,we use the definition $f(u) = 3 + u$. Thus,$\lim_{x 	o 0^-} f(f(x)) = \lim_{u 	o 2^+} (3 + u) = 3 + 2 = 5$.For the $RHL$ $(x 	o 0^+)$: As $x 	o 0^+$,$f(x) = 3 + x$. Since $x$ is slightly greater than $0$,$f(x)$ is slightly greater than $3$. Let $u = f(x)$. As $x 	o 0^+$,$u 	o 3^+$. Since $u > 0$,we use the definition $f(u) = 3 + u$. Thus,$\lim_{x 	o 0^+} f(f(x)) = \lim_{u 	o 3^+} (3 + u) = 3 + 3 = 6$.Since the $LHL$ $(5)$ is not equal to the $RHL$ $(6)$,the limit does not exist.

If $f(x) = \begin{cases} 3 + x; & x \geqslant 0 \\ 2 - 3x; & x < 0 \end{cases}$,then $\lim_{x \to 0} f(f(x))$ is equal to -

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