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  • Try this! - Limit , continuity and differentiability - JEE Main-4

\lim_{x\rightarrow 0} \frac{(27+x)^{1/3}-3}{9-(27+x)^{2/3}}

equals 

  • Option 1)

    1/3

  • Option 2)

    -1/3

  • Option 3)

    -1/6

  • Option 4)

    1/6

 

Answers (2)

best_answer

As we have learned

Limit of product / quotient -

Limit of product is the product of individual limits such that

\lim_{x\rightarrow a}{f(x).g(x)}

=\lim_{x\rightarrow a}{f(x).\lim_{x\rightarrow a}g(x)

also\:\lim_{x\rightarrow a}{kf(x)}

=k\lim_{x\rightarrow a}{f(x)}

\lim_{x\rightarrow a}\frac{f(x)}{g(x)}=\frac{\lim_{x\rightarrow a}f(x)}{\lim_{x\rightarrow a}g(x)}

 

-

 

 using approximation (1+x)^{n}\approx 1+nx

\Rightarrow \lim_{x\rightarrow 0} \frac{3\left [ 1+\frac{1}{3}\times \frac{x}{27}-1 \right ]}{9\left [ 1-1-\frac{x}{27}\times \frac{2}{3} \right ]}

= -1/6

 

 

 


Option 1)

1/3

This is incorrect 

Option 2)

-1/3

This is incorrect 

Option 3)

-1/6

This is correct 

Option 4)

1/6

This is incorrect 

Posted by

Himanshu

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