If gx - x-x- for x - 2 then limx- 2-gx - g2x - 2 -

The correct option is D 12 g(x) = x2 for 2 lt;x lt;3 ∴ #160;lim_x→ 2^+g(x) g(2)x 2 = lim_x→ 2^+x2 1x 2 = 12 ...

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Factorization Method Form to Remove Indeterminate Form

If gx = x[x...

Question

If $g (x) = \frac{x}{[x]}$ for $x > 2$ then $lim x \to 2^{+} \frac{g (x) - g (2)}{x - 2} =$

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Similar questions

g (x) = \frac{x}{[\begin{matrix} x \end{matrix}]}

x > 2

l i m x \to 2^{+} \frac{g (x) - g (2)}{x - 2} =

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) = {\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}

h (x) = | x |

lim x \to 0 f (g (h (x)))

f (x) = {\begin{matrix} sin x; x \neq n π, n = 0, \pm 1, \pm 2, \pm 3..... 2; o t h e r w i s e \end{matrix}

g (x) = {\begin{matrix} x^{2} + 1; x \neq 0 4; x = 0 \end{matrix} .

{lim}_{x \to 0} g (f (x))

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