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  • Can someone explain - Limit , continuity and differentiability - JEE Main-2

\lim_{x\rightarrow \infty }\left ( \frac{x^{2}+5x+3}{x^{2}+x+3} \right )^{\frac{1}{x}}

  • Option 1)

    e^{4}

  • Option 2)

    e^{2}

  • Option 3)

    e^{3}

  • Option 4)

    1

 

Answers (1)

best_answer

As we learnt in 

1 to the power of infinity Form -

Let\:\:\;\lim_{x\rightarrow a}f(x)^{g(x)}\:\;\:where

f(a)=1\:\:\:and \;\:\:g(a)=\infty

Then\:\:\:\:e^\lim_{x\rightarrow a}(f(x)-1)g(x)

-

 

 

\lim_{n \to \infty } \left ( \frac{x^{2}+5x+3}{x^{2}+x+3} \right )^{\frac{1}{x}}

\1^{\infty } \: form

\lim_{n \to \infty } \left ( \frac{x^{2}+5x+3}{x^{2}+x+3}-1 \right )\times \frac{1}{x}

\lim_{n \to \infty } \left ( \frac{x^{2}+5x+3-x^{2}-x-3}{x^{2}+x+3}\right )\times \frac{1}{x}

\lim_{n \to \infty } \frac{4x}{x^{2}+x+3}\times \frac{1}{x}= \frac{4}{\infty }=0

\therefore \: e^{0}=1


Option 1)

e^{4}

This option is incorrect

Option 2)

e^{2}

This option is incorrect

Option 3)

e^{3}

This option is incorrect

Option 4)

1

This option is correct

Posted by

divya.saini

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