Questions & Answers

Question

Answers

A. –2

B. 1

C. 0

D. None of these

Answer
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Intermediate forms are those forms which cannot be calculated by directly putting the limits in the given equation.

Like \[\dfrac{0}{0}\] or \[\dfrac{\infty }{\infty }\] are some of the intermediate forms.

So, to check whether the given equation is intermediate form at the given limit we had to put the limit in the given equation.

So, now putting the limits.

\[ \Rightarrow \mathop {\lim }\limits_{x \to \pi /4} \dfrac{{{{\left( {\tan \dfrac{\pi }{4}} \right)}^{\tan \dfrac{\pi }{4}}} - \tan \dfrac{\pi }{4}}}{{\ln \left( {\tan \dfrac{\pi }{4}} \right) - \tan \dfrac{\pi }{4} + 1}}\] (1)

Now as we know that \[\tan \dfrac{\pi }{4} = 1\]. So, above equation becomes,

\[ \Rightarrow \mathop {\lim }\limits_{x \to \pi /4} \dfrac{{{{\left( 1 \right)}^1} - 1}}{{\ln \left( 1 \right) - 1 + 1}} = \dfrac{0}{0}\] (because \[\ln \left( 1 \right) = 0\])

Now as we can see that the given equation is of intermediate form i.e. \[\dfrac{0}{0}\] form. So, we can directly LH Rule.

According to LH Rule we had to apply differentiation on numerator and denominator separately, still it holds indefinite form or intermediate form after its application.

So, now differentiate numerator and denominator of equation 1 with respect to x separately.

\[ \Rightarrow \mathop {\lim }\limits_{x \to \pi /4} \dfrac{{\dfrac{{d\left( {{{\left( {\tan x} \right)}^{\tan x}}} \right)}}{{dx}} - \dfrac{{d\left( {\tan x} \right)}}{{dx}}}}{{\dfrac{{d\left( {\ln \left( {\tan x} \right)} \right)}}{{dx}} - \dfrac{{d\left( {\tan x} \right)}}{{dx}} + \dfrac{{d\left( 1 \right)}}{{dx}}}}\] (2)

Let \[y = {\left( {\tan x} \right)^{\tan x}}\]

So, now taking log both sides of the above equation.

\[ \Rightarrow \log \left( y \right) = \tan x\left( {\log \left( {\tan x} \right)} \right)\]

Now let us find the derivative of the above equation with respect to x by using product rule.

\[ \Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = {\sec ^2}x\left( {\log \left( {\tan x} \right)} \right) + \tan x\left( {\dfrac{1}{{\tan x}}{{\sec }^2}x} \right)\]

Putting the value of y in the above equation and then multiplying both sides of the above equation by y.

\[ \Rightarrow \dfrac{{d\left( {\tan {x^{\tan x}}} \right)}}{{dx}} = \tan {x^{\tan x}}{\sec ^2}x\left( {\log \left( {\tan x} \right) + 1} \right)\]

Now equation 2 becomes,

\[ \Rightarrow \mathop {\lim }\limits_{x \to \pi /4} \dfrac{{\tan {x^{\tan x}}{{\sec }^2}x\left( {\log \left( {\tan x} \right) + 1} \right) - {{\sec }^2}x}}{{\dfrac{1}{{\tan x}} \times {{\sec }^2}x - {{\sec }^2}x + 0}}\]

Now take \[{\sec ^2}x\] common from the numerator and the denominator of the above equation and then cancel.

\[ \Rightarrow \mathop {\lim }\limits_{x \to \pi /4} \dfrac{{\tan {x^{\tan x}}\left( {\log \left( {\tan x} \right) + 1} \right) - 1}}{{\dfrac{1}{{\tan x}} - 1}}\]

Now let us put the limit in above equation to check whether it is in intermediate form or not

\[ \Rightarrow \mathop {\lim }\limits_{x \to \pi /4} \dfrac{{{{\left( {\tan \dfrac{\pi }{4}} \right)}^{\tan \dfrac{\pi }{4}}}\left( {\log \left( {\tan \dfrac{\pi }{4}} \right) + 1} \right) - 1}}{{\cot \dfrac{\pi }{4} - 1}} = \dfrac{0}{0}\] form

So, now let us apply the LH Rule again.

So, differentiate numerator and denominator of equation 3 with respect to x separately.

\[ \Rightarrow \mathop {\lim }\limits_{x \to \pi /4} \dfrac{{\dfrac{{d\left( {\tan {x^{\tan x}}\log \left( {\tan x} \right)} \right)}}{{dx}} + \dfrac{{d\left( {\tan {x^{\tan x}}} \right)}}{{dx}} - \dfrac{{d\left( 1 \right)}}{{dx}}}}{{\dfrac{{d\left( {\cot x} \right)}}{{dx}} - \dfrac{{d\left( 1 \right)}}{{dx}}}}\]

\[ \Rightarrow \mathop {\lim }\limits_{x \to \pi /4} \dfrac{{\left( {\tan {x^{\tan x}}{{\sec }^2}x\left( {\log \left( {\tan x} \right) + 1} \right)\log \left( {\tan x} \right) + \tan {x^{\tan x}} \times \dfrac{1}{{\tan x}} \times {{\sec }^2}x + \tan {x^{\tan x}}{{\sec }^2}x\left( {\log \left( {\tan x} \right) + 1} \right)} \right)}}{{ - \cos e{c^2}x}}\]

Now put the limits \[x \to \dfrac{\pi }{4}\] in the above equation to check whether it is in intermediate form or not.

\[ \Rightarrow \mathop {\lim }\limits_{x \to \pi /4} \dfrac{{{{\left( {\tan \dfrac{\pi }{4}} \right)}^{\tan \dfrac{\pi }{4}}}{{\sec }^2}\dfrac{\pi }{4}\left( {\left( {\left( {\log \left( {\tan \dfrac{\pi }{4}} \right) + 1} \right)\log \left( {\tan \dfrac{\pi }{4}} \right) + \left( {\dfrac{1}{{\tan \dfrac{\pi }{4}}}} \right) + \left( {\log \left( {\tan \dfrac{\pi }{4}} \right) + 1} \right)} \right)} \right)}}{{ - \cos e{c^2}\dfrac{\pi }{4}}}\]

\[ \Rightarrow \mathop {\lim }\limits_{x \to \pi /4} \dfrac{{\left( 1 \right) \times 2\left( {\left( {\left( {\log \left( 1 \right) + 1} \right)\log \left( 1 \right) + \left( {\dfrac{1}{1}} \right) + \left( {\log \left( {\tan 1} \right) + 1} \right)} \right)} \right)}}{{ - 2}}\]

Now as we know that \[\log \left( 1 \right) = 0\]. So, above equation becomes,

\[ \Rightarrow \mathop {\lim }\limits_{x \to \pi /4} \dfrac{{2\left( {0 + 1 + 1} \right)}}{{ - 2}} = \dfrac{4}{{ - 2}} = - 2\]

So, the value of \[\mathop {\lim }\limits_{x \to \pi /4} \dfrac{{{{\left( {\tan x} \right)}^{\tan x}} - \tan x}}{{\ln \left( {\tan x} \right) - \tan x + 1}}\] will be equal to \[ - 2\].

Hence, the correct option will be A.

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