If fx - x - 1- x - 1 2x2 - 2- x 0 -x2 - 1- x - 0- and hx - -x- then limx - 0 fg h x is

The correct option is A 0 lim_x → 0^+ f(g(h(x))) = f(g(0^+)) = f(1^+) = 0 lim_x → 0^ f(g(h(x))) = f(g(0^+)) = f(1^+) = 0 Therefore, lim_x → 0 f( ...

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Substitution Method to Remove Indeterminate Form

If fx = x -...

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If $f (x) = {\begin{matrix} x - 1, & x \geq 1 2 x^{2} - 2, & x < 1 \end{matrix}, g (x) = {\begin{matrix} x + 1, & x > 0 - x^{2} + 1, & x \leq 0 \end{matrix}$ , and $h (x) = | x |$ , then $lim x \to 0 f (g (h (x)))$ is

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f (x) = {\begin{matrix} x - 1, x \geq 1, 2 x^{2} - 2, x < 1 \end{matrix} g (x) = {\begin{matrix} x + 1, x > 0 - x^{2} + 1, x \leq 0 \end{matrix} and h (x) = | x |, then lim x \to 0 f (g (h (x)))

f (x) = [\frac{x}{[x]}], g (x) = [x [\frac{1}{x}]]

h (x) = [\frac{[x]}{x}],

lim x \to 2^{-} f (x) + lim x \to {\frac{1}{2}}^{+} g (x) + lim x \to 2^{+} h (x) =

f (x) = \frac{2}{x - 3}

g (x) = \frac{x - 3}{x + 4}

h (x) = - \frac{2 (2 x + 1)}{x^{2} + x - 12}

lim x \to 3 [f (x) + g (x) + h (x)]

f (x) = x^{2} + \frac{1}{x^{2}} and g (x) = x - \frac{1}{x}, x ϵ R - {- 1, 0, 1}, and h (x) = \frac{f (x)}{g (x)}

h (x)

f (x) = \frac{2}{x - 3}, g (x) = \frac{x - 3}{x + 4}

h (x) = \frac{- 2 (2 x + 1)}{x^{2} + x - 12}

lim x \to 3 [f (x) + g (x) + h (x)]

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