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The given function is $f(x) = x|x|$. We need to check its differentiability at $x=0$. First, we express the function in piecewise form:
$$f(x) = \begin{cases} x^2, & \text{if } x \geq 0 \\ -x^2, & \text{if } x < 0 \end{cases}$$
The left-hand derivative at $x=0$ is given by:
$$LHD = \lim_{h \to 0^-} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^-} \frac{f(h) - f(0)}{h}$$
Since $h \to 0^-$, $h < 0$, so $f(h) = -h^2$. Also, $f(0) = 0$. Therefore,
$$LHD = \lim_{h \to 0^-} \frac{-h^2 - 0}{h} = \lim_{h \to 0^-} -h = 0$$
The right-hand derivative at $x=0$ is given by:
$$RHD = \lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^+} \frac{f(h) - f(0)}{h}$$
Since $h \to 0^+$, $h > 0$, so $f(h) = h^2$. Also, $f(0) = 0$. Therefore,
$$RHD = \lim_{h \to 0^+} \frac{h^2 - 0}{h} = \lim_{h \to 0^+} h = 0$$
We have $LHD = 0$ and $RHD = 0$. Since $LHD = RHD$, the function is differentiable at $x=0$.
Final Answer: The function is differentiable at x=0.
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