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Which of the following is true for the signum function \operatorname{sgn}(x) ? 

Option: 1

\lim _{x \rightarrow 0^{+}} \operatorname{sgn}(x)=\lim _{x \rightarrow 0} \operatorname{sgn}(x)=0


Option: 2

\lim _{x \rightarrow 0^{+}} \operatorname{sgn}(x)=\lim _{x \rightarrow 0} \operatorname{sgn}(x)=1


Option: 3

\lim _{x \rightarrow 0^{+}} \operatorname{sgn}(x)+\lim _{x \rightarrow 0} \operatorname{sgn}(x)=0


Option: 4

\lim _{x \rightarrow 0^{+}} \operatorname{sgn}(x)-\lim _{x \rightarrow 0} \operatorname{sgn}(x)=0


Answers (1)

best_answer

Note that the following essential points.

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

  • The Left Hand Limit (LHL) of the function f\left ( x \right ) at x=a is \lim _{x \rightarrow a^{-}} f(x)=\lim _{h \rightarrow 0} f(a-h) .

  • Here, h is positive and infinitely small.

  • The limit exists only when \lim _{x \rightarrow a+} f(x)=\lim _{x \rightarrow a-} f(x)

  • The signum function returns the sign of the variable in the following way.

  1. When x>0 , its value \operatorname{sgn}(x)=1 .

  2. When x<0 , its value \operatorname{sgn}(x)=-1 .

  3. When x=0 , its value \operatorname{sgn}(x)=0 .

Since, the domain of the signum function is and its range is the set \{-1,0,1\} , its graphical representation is 

The Left Hand limit of   \lim _{x \rightarrow 0} \operatorname{sgn}(x)   is

\begin{aligned} & \lim _{x \rightarrow 0^{-}} \operatorname{sgn}(x) \\ & =1 \end{aligned}

The Right Hand limit of  \lim _{x \rightarrow 0} \operatorname{sgn}(x)   is

\begin{aligned} & \lim _{x \rightarrow 0^{-}} \operatorname{sgn}(x) \\ & =-1 \end{aligned}

Now, deduce the following 

\begin{aligned} & \cdot \lim _{x \rightarrow 0^{+}} \operatorname{sgn}(x)+\lim _{x \rightarrow 0^{-}} \operatorname{sgn}(x)=1-1=0 \\ &\cdot \lim _{x \rightarrow 0^{+}} \operatorname{sgn}(x)-\lim _{x \rightarrow 0^{-}} \operatorname{sgn}(x)=1-(-1)=2 \end{aligned} 

 

 

 

  

Posted by

HARSH KANKARIA

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