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What is the Left Hand limit of the function \lim _{x \rightarrow-1} \frac{[x]}{x^2+5}, where  \left [ \ \right ]  indicates the Floor Function?

 

Option: 1

\begin{aligned} & \frac{1}{6} \\ \end{aligned}


Option: 2

\begin{aligned} & -\frac{1}{6} \\ \end{aligned}


Option: 3

\begin{aligned} & -\frac{1}{4} \\ \end{aligned}


Option: 4

\begin{aligned} & -\frac{1}{3} \end{aligned}


Answers (1)

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Note that the following essential points.

  • The Right Hand limit of the function at is f(x) \text { at } x=a \text { is } \lim _{x \rightarrow a^{+}} f(x)=\lim _{h \rightarrow 0} f(a+h).

  • The Left Hand limit of the function at is f(x) \text { at } x=a \text { is } \lim _{x \rightarrow a^{+}} f(x)=\lim _{h \rightarrow 0} f(a-h) .

  • Here, h is positive and infinitely small.

  • The Floor Function indicates the greatest integer function [GIF] denoted mathematically as  [p] , for only real values. This function  [p]  rounds downs the real number having any fractional or decimal part (if any) to the nearest integral value less than the indicated number.

For  h \to 0^{-}   the GIF [-1-h]=-2

The Left Hand limit of the function \lim _{x \rightarrow-1} \frac{[x]}{x^2+5}   is

\begin{aligned} & \lim _{x \rightarrow(-1)^{-}} \frac{[x]}{x^2+5} \\ & =\lim _{h \rightarrow 0^{-}} \frac{[-1-h]}{(-1+h)^2+5} \\ & =\lim _{h \rightarrow 0^{-}} \frac{[-1-h]}{1-2 h+h^2+5} \\ & =\frac{-2}{1-0+0+5} \\ & =-\frac{2}{6} \\ & =-\frac{1}{3} \end{aligned}

 

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jitender.kumar

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