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  • Please help! - Limit , continuity and differentiability - JEE Main-5

\lim_{x\rightarrow 1} \frac{x^{3}-xlnx+lnx-1}{x^{2}-1} equals

  • Option 1)

    3/2

  • Option 2)

    3

  • Option 3)

    6

  • Option 4)

    12

 

Answers (1)

best_answer

As we have learned

Method of factorisation -

 Indeterminate\:form\:of\:\frac{0}{0}\:and \:\frac{\infty }{\infty }

We remove the denominator factor which it makes zero.


\lim_{x\rightarrow 1}\:\frac{x^{2}-1}{x-1}=\lim_{x\rightarrow 1}=\frac{(x-1)(x+1)}{(x-1)}=1+1=2

 

- wherein

\frac{0}{finite}=0


\frac{finite}{0}=\infty

 

 

\lim_{x\rightarrow 1}\frac{x^{3}-xlnx+lnx-1}{x^{2}-1}=\lim_{x\rightarrow 1}\frac{(x^{3}-1)(lnx)(x-1)}{x^{2}-1}

\lim_{x\rightarrow 1}\frac{(x-1)(x^{2}+x+1-lnx)}{(x-1)(x+1)}=\lim_{x\rightarrow 1}\frac{x^{2}+x+1-lnx}{x+1}=3/2

 

 

 

 


Option 1)

3/2

Option 2)

3

Option 3)

6

Option 4)

12

Posted by

Himanshu

View full answer

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