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  • Need clarity, kindly explain! - Limit , continuity and differentiability - JEE Main-2

If x approachs 2, then approximate value of \frac{x^{2}-4}{x-2} is

  • Option 1)

    1

  • Option 2)

    2

  • Option 3)

    3

  • Option 4)

    4

 

Answers (1)

best_answer

As we have learned

Condition on Limits -

Limit does not give actual value.It gives approximate value.

- wherein

f(x)=\frac{x^{2}+x-2}{x-1}


x is not defined at  x = 1 but for  x > 1  &  x < 1  it gives approximate values.

 

\frac{x^{2}-4}{x-2}=\frac{(x-2)(x+2)}{(x-2)} 

When  x approaches 2, \frac{x^{2}-4}{x-2} simplifies to x+2 

\therefore x+2 approaches to 4

 

 

 


Option 1)

1

Option 2)

2

Option 3)

3

Option 4)

4

Posted by

Himanshu

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