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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} \\
 $

Answer Verified Verified
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
\[\mathop {\lim }\limits_{x \to 0} {(f(x))^{g(x)}} = {e^{\mathop {\lim }\limits_{x \to 0 + } \dfrac{{(f(x) - 1)}}{{g(x)}}}}\]

Complete step-by-step answer:
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 $10$.