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  • Tell me? - Limit , continuity and differentiability - JEE Main-3

\lim_{x\rightarrow -2}\frac{x^{5}+32}{x^{3}+8} equals 

  • Option 1)

    20/3

  • Option 2)

    10/3

  • Option 3)

    5/3

  • Option 4)

    4/3

 

Answers (1)

best_answer

As we have learned

Evaluation of limits using standard results -

\lim_{x\rightarrow a}\:\frac{{x^{n}}-{a^{n}}}{x-a}=n a^{n-1}

- wherein

\because \:{x^{n}}-{a^{n}}=(x-a)({x}^{n-1}+{x}^{n-2}a +{x}^{n-3}a^{2}+. ......{a}^{n-1})

 

 \lim_{x\rightarrow -2}\frac{x^{5}+32}{x^{3}+8}\lim_{x\rightarrow -2}\frac{x^{5}-(-2)^{5}}{x^{3}-(-2)^{3}}  dividing numerator and denominator by x+2 

\lim_{x\rightarrow -2}\frac{\frac{x^{5}-(-2^{5})}{x-(-2)}}{\frac{x^{3}-(-2)^{3}}{x-(-2)}}= \frac{5(-2)^{4}}{3(-2)^{2}}=20/3

 

 

 

 


Option 1)

20/3

Option 2)

10/3

Option 3)

5/3

Option 4)

4/3

Posted by

Himanshu

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