Question
Answers

The value of \[\mathop {\lim }\limits_{x \to \pi /4} \dfrac{{{{\left( {\tan x} \right)}^{\tan x}} - \tan x}}{{\ln \left( {\tan x} \right) - \tan x + 1}}\] is
A. –2
B. 1
C. 0
D. None of these

Answer Verified Verified
Hint: Check whether the given equation is \[\dfrac{0}{0}\] or \[\dfrac{\infty }{\infty }\] form at the given limit \[x \to \dfrac{\pi }{4}\] and if it is \[\dfrac{0}{0}\] or \[\dfrac{\infty }{\infty }\] form then we can directly apply L Hospital Rule which states that differentiate numerator and denominator separately and apply this rule as long asit holds any indefinite form every time after its application.

Complete step-by-step answer:
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.

Note:Whenever we come up with this type of problem then first we had to put the given limit (here \[x \to \dfrac{\pi }{4}\]) and if the result is not in intermediate form then that will be the required answer and if it is in intermediate form then we had to apply LH rule till it holds any indefinite form and check the value of equation after every limit and note that when we differentiate numerator and denominator then after that before putting limits wen should cancel the common terms from numerator and denominator otherwise it can give the intermediate form. This will be the easiest and efficient way to find the solution of the problem.


Bookmark added to your notes.
View Notes
×