lim_{x rightarrow -2^{+}} ([x]^2 - [x] - 2) + | vedclass

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Question

(A) પ્રથમ લક્ષ માટે: $\lim_{x \rightarrow -2^{+}} [x] = -2$. તેથી,$\lim_{x \rightarrow -2^{+}} ([x]^2 - [x] - 2) = (-2)^2 - (-2) - 2 = 4 + 2 - 2 = 4$.

Vedclass · Accepted Answer

$39$

Vedclass · Answer

(A) પ્રથમ લક્ષ માટે: $\lim_{x \rightarrow -2^{+}} [x] = -2$. તેથી,$\lim_{x \rightarrow -2^{+}} ([x]^2 - [x] - 2) = (-2)^2 - (-2) - 2 = 4 + 2 - 2 = 4$. બીજા લક્ષ માટે: $\lim_{x \rightarrow -3^{-}} [x] = -4$. તેથી,$\lim_{x \rightarrow -3^{-}} ([x]^2 - 4[x] + 3) = (-4)^2 - 4(-4) + 3 = 16 + 16 + 3 = 35$. બંને પરિણામોનો સરવાળો: $4 + 35 = 39$.

$\lim_{x \rightarrow -2^{+}} ([x]^2 - [x] - 2) + \lim_{x \rightarrow -3^{-}} ([x]^2 - 4[x] + 3) =$

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