# Let $p = \mathop {\lim }\limits_{x \to 0 + } {(1 + {\tan ^2}\sqrt x )^{\dfrac{1}{{2x}}}}$then $\log p$ is equal to:$a)\,2 \\ b)\,1 \\ c)\,\dfrac{1}{2} \\ d)\,\dfrac{1}{4} \\$

Verified
142.2k+ views
Hint:By observing the equation we can see that the intermediate form ${1^\infty }$. So we solve it according to this
$\mathop {\lim }\limits_{x \to 0} {(f(x))^{g(x)}}$where for $x \to 0\,f(x) \to 1$and for $x \to 0\,\,\,g(x) \to \infty$. Then
$\mathop {\lim }\limits_{x \to 0} {(f(x))^{g(x)}} = {e^{\mathop {\lim }\limits_{x \to 0 + } \dfrac{{(f(x) - 1)}}{{g(x)}}}}$

Given $p = \mathop {\lim }\limits_{x \to 0 + } {(1 + {\tan ^2}\sqrt x )^{\dfrac{1}{{2x}}}}$ $\to (1)$
We can see that for $x \to {0^ + },1 + {\tan ^2}\sqrt x \to 1$and for $x \to {0^ + },\dfrac{1}{{2x}} \to \infty$
So for $x \to {0^ + },{(1 + {\tan ^2}\sqrt x )^{\dfrac{1}{{2x}}}} \to {1^\infty }$
So it is an intermediate form ${1^\infty }$.
And we know for ${1^\infty }$intermediate form $\mathop {\lim }\limits_{x \to 0} {(f(x))^{g(x)}}$where for $x \to 0\,f(x) \to 1$and for $x \to 0\,\,\,g(x) \to \infty$.
$\mathop {\lim }\limits_{x \to 0} {(f(x))^{g(x)}} = {e^{\mathop {\lim }\limits_{x \to 0 + } \dfrac{{(f(x) - 1)}}{{g(x)}}}}$ $\to (2)$
Now using (2) in (1) we get,
$p = \mathop {\lim }\limits_{x \to 0 + } {(1 + {\tan ^2}\sqrt x )^{\dfrac{1}{{2x}}}} = {e^{\mathop {\lim }\limits_{x \to 0} \dfrac{1}{{2x}}(1 + {{\tan }^2}\sqrt x - 1)}} \\ p = {e^{\mathop {\lim }\limits_{x \to 0} \dfrac{{{{\tan }^2}\sqrt x }}{{2x}}}} \\ p = {e^{\mathop {\lim }\limits_{x \to 0} \dfrac{{{{\tan }^2}\sqrt x }}{{2{{\sqrt x }^2}}}}} \\ p = {e^{\mathop {\lim }\limits_{x \to 0} {{\left( {\dfrac{{\tan \sqrt x }}{{\sqrt x }}} \right)}^2} \times \dfrac{1}{2}}} \\$
Now using $\mathop {\lim }\limits_{x \to 0 + } {\left( {\dfrac{{\tan \sqrt x }}{{\sqrt x }}} \right)^2} = 1$. We get,
$p = {e^{\dfrac{1}{2}(1)}} \\ p = {e^{\dfrac{1}{2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to (3) \\$
Now we need to find the value of $\log p$, according to the question.
So taking log on both sides in (3), we get
$\log p = \log {e^{\dfrac{1}{2}}}$
Now using $\log {e^b} = b\log e = b(1) = b$
$\log p = \dfrac{1}{2}\log e = \dfrac{1}{2}$
So $\log p = \dfrac{1}{2}$

So, the correct answer is “Option C”.

Note:Using $\tan x = x + \dfrac{{{x^3}}}{3} + \dfrac{{2{x^5}}}{{15}} + .............$.
We can see that
$\mathop {\lim }\limits_{x \to 0 + } {\left( {\dfrac{{\tan \sqrt x }}{{\sqrt x }}} \right)^2} = \mathop {\lim }\limits_{x \to 0 + } \left( {\dfrac{{\sqrt x + \dfrac{{{{\sqrt x }^3}}}{3} + \dfrac{{2{{\sqrt x }^5}}}{{15}} + ...........}}{{\sqrt x }}} \right) \\ = \mathop {\lim }\limits_{x \to 0 + } \left( {1 + \dfrac{{{{\sqrt x }^2}}}{3} + \dfrac{{2{{\sqrt x }^4}}}{{15}} + ..........} \right) \\ = 1 \\$
This point is important to solve this question. And we take $\log$ with base $e$ and then ${\log _e}e = 1$. And will not take $\log$ with base $10$.