Let p= \lim_{x\rightarrow 0+}\left ( 1+\tan ^{2} \sqrt{x}\right )^{\frac{1}{2x}}then  log p
is equal to :

  • Option 1)

    2

  • Option 2)

    1

  • Option 3)

    \frac{1}{2}

  • Option 4)

    \frac{1}{4}

 

Answers (1)

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)

-

 

P=\lim_{x\rightarrow 0^{+}}\left ( 1+tan^{2}\sqrt{x} \right )^\frac{1}{2x}

It is 1^{\infty} form

\therefore P=\lim_{x\rightarrow 0^{+}}\left(1+tan^{2}\sqrt{x}-1 \right )\times \frac{1}{2x}

\therefore \lim_{x\rightarrow 0^{+}}\:\frac{tan^{2}\sqrt{x}}{2x}=\frac{1}{2} 

\therefore \:P=e^\frac{1}{2}

logP=\frac{1}{2}


Option 1)

2

This option is incorrect.

Option 2)

1

This option is incorrect.

Option 3)

\frac{1}{2}

This option is correct.

Option 4)

\frac{1}{4}

This option is incorrect.

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