At the point x=1,the function f(x) = begin{ca | vedclass

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(B) $LHL$ $= \lim_{x \rightarrow 1^-} f(x) = \lim_{x \rightarrow 1} (x-1) = 0$ $RHL$ $= \lim_{x \rightarrow 1^+} f(x) = \lim_{x \rightarrow 1} (x^3-1)

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continuous and not differentiable

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(B) $LHL$ $= \lim_{x ightarrow 1^-} f(x) = \lim_{x ightarrow 1} (x-1) = 0$ $RHL$ $= \lim_{x ightarrow 1^+} f(x) = \lim_{x ightarrow 1} (x^3-1) = 0$ Also,$f(1) = 1-1 = 0$ Since $LHL$ $=$ $RHL$ $= f(1)$,$f$ is continuous at $x=1$. Now,$Lf'(1) = \lim_{h ightarrow 0} \frac{f(1-h)-f(1)}{-h} = \lim_{h ightarrow 0} \frac{(1-h)-1-0}{-h} = \lim_{h ightarrow 0} \frac{-h}{-h} = 1$ And $Rf'(1) = \lim_{h ightarrow 0} \frac{f(1+h)-f(1)}{h} = \lim_{h ightarrow 0} \frac{(1+h)^3-1-0}{h} = \lim_{h ightarrow 0} \frac{1+h^3+3h+3h^2-1}{h} = \lim_{h ightarrow 0} (h^2+3h+3) = 3$ Since $Lf'(1) 
eq Rf'(1)$,$f(x)$ is not differentiable at $x=1$.

$(a)$ $x\|x\|$	$(i)$ continuous in $(-1, 1)$
$(b)$ $\sqrt{\|x\|}$	$(ii)$ differentiable in $(-1, 1)$
$(c)$ $x+[x]$	$(iii)$ strictly increasing in $(-1, 1)$
$(d)$ $\|x-1\|+\|x+1\|$	$(iv)$ not differentiable at,at least one point in $(-1, 1)$

At the point $x=1$,the function $f(x) = \begin{cases} x^3-1, & 1 < x < \infty \\ x-1, & -\infty < x \leq 1 \end{cases}$ is

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