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  • Confused! kindly explain, - Limit , continuity and differentiability - JEE Main-4

If x approachs 3, then  \frac{x^{2}-5x+6}{x^{2}-4x+3}  has approximate value

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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}-5x+6}{x^{2}-4x+3}= \frac{(x-2)(x-3)}{(x-1)(x-3)}

When x approaches 3, \frac{x^{2}-5x+6}{x^{2}-4x+3} simplifies to \frac{x-2}{x-1} 

\therefore \frac{x-2}{x-1}   approaches \frac{3-2}{3-1} =1/2

 

 

 


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