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\lim_{x\rightarrow 0^{+}}   sgn(x) equals 

  • Option 1)

    -1

  • Option 2)

    0

  • Option 3)

    1

  • Option 4)

    2

 

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As we have learned

Right hand limit -

The right hand limit of  f(x) as 'x' tends to 'a' exists and is equal to  l1  if as  'x'  approaches 'a' through values greater than 'a'.
 

so\:\lim_{x\rightarrow a^{+}}f(x)=l_{1} 

- wherein

where  a+ means  a+h  &  h → 0  therefore f(a+h)

 

 We know , sgn (x) = \left \{ -1 ;x< 0 \right.

                                  \left \{ 0 ;x= 0 \right.

                                   \left \{ 1 ;x> 0 \right.

In \lim_{x\rightarrow 0^{+}} sgn (x) , since x> 0 so its limit will be 1

 

 

 

 


Option 1)

-1

Option 2)

0

Option 3)

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Option 4)

2

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