Calculate the following limit: \[\displaystyle \lim_{x \to \dfrac{\pi }{4}}\dfrac{{{\cot }^{3}}x-\tan x}{\cos \left( x+\dfrac{\pi }{4} \right)}\]
A). $4$
B). $8\sqrt{2}$
C). $8$
D). $4\sqrt{2}$

Hint: We will first find out whether the given function is in the form of $\left( \dfrac{0}{0} \right)$ or $\left( \dfrac{\infty }{\infty } \right)$ by putting x as $\dfrac{\pi }{4}$in the function, We will then apply the L-Hospital Rule to find the definite value of the limit. After differentiating the numerator and the denominator, we will again put x as $\dfrac{\pi }{4}$ in the function to get the final limit

Complete step-by-step answer:
To find the limit of the given function: \[\displaystyle \lim_{x \to \dfrac{\pi }{4}}\dfrac{{{\cot }^{3}}x-\tan x}{\cos \left( x+\dfrac{\pi }{4} \right)}\]
As x is approaching $\dfrac{\pi }{4}$ as shown in the function, we will put as x as $\dfrac{\pi }{4}$ and see the value that comes:
Let’s take $f\left( x \right)=\dfrac{{{\cot }^{3}}x-\tan x}{\cos \left( x+\dfrac{\pi }{4} \right)}$, now when we put x as
$f\left( \dfrac{\pi }{4} \right)=\dfrac{{{\cot }^{3}}\dfrac{\pi }{4}-\tan \dfrac{\pi }{4}}{\cos \left( \dfrac{\pi }{4}+\dfrac{\pi }{4} \right)}\Rightarrow f\left( \dfrac{\pi }{4} \right)=\dfrac{{{\cot }^{3}}\dfrac{\pi }{4}-\tan \dfrac{\pi }{4}}{\cos \left( \dfrac{\pi }{2} \right)}............\text{ Equation 1}\text{.}$
Now, we know that:
$\begin{align}
  & \cot \theta =\dfrac{\cos \theta }{\sin \theta }\Rightarrow \cot \dfrac{\pi }{4}=\dfrac{\cos \left( \dfrac{\pi }{4} \right)}{\sin \left( \dfrac{\pi }{4} \right)}\text{ }...........\text{Equation 2}\text{.} \\
 & \\
\end{align}$
We already know that the value of $\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}$ and $\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}$ , so we will putting these values in Equation 2 in order to find value of $\cot \dfrac{\pi }{4}$:
$\cot \dfrac{\pi }{4}=\dfrac{\cos \left( \dfrac{\pi }{4} \right)}{\sin \left( \dfrac{\pi }{4} \right)}\Rightarrow \cot \dfrac{\pi }{4}=\dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{1}{\sqrt{2}}}$
After cancelling out $\dfrac{1}{\sqrt{2}}$ , we get $\cot \dfrac{\pi }{4}=1$.
Similarly we will find out the value of $\tan \dfrac{\pi }{4}$ :
$\tan \theta =\dfrac{\sin \theta }{\cos \theta }\Rightarrow \tan \dfrac{\pi }{4}=\dfrac{\sin \left( \dfrac{\pi }{4} \right)}{\cos \left( \dfrac{\pi }{4} \right)}\text{ }...........\text{Equation 3}\text{.}$
We already know that the value of $\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}$ and $\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}$ , so we will putting these values in Equation 3 in order to find the value of $\tan \dfrac{\pi }{4}$:
$\tan \dfrac{\pi }{4}=\dfrac{\sin \left( \dfrac{\pi }{4} \right)}{\cos \left( \dfrac{\pi }{4} \right)}\Rightarrow \tan \dfrac{\pi }{4}=\dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{1}{\sqrt{2}}}$
After cancelling out $\dfrac{1}{\sqrt{2}}$ , we get $\tan \dfrac{\pi }{4}=1$.
Putting the values of $\tan \dfrac{\pi }{4}$ and $\cot \dfrac{\pi }{4}$ in Equation 1.
We have: $f\left( \dfrac{\pi }{4} \right)=\dfrac{{{\cot }^{3}}\dfrac{\pi }{4}-\tan \dfrac{\pi }{4}}{\cos \left( \dfrac{\pi }{2} \right)}$ , We found out that $\cot \dfrac{\pi }{4}=1$ and $\tan \dfrac{\pi }{4}=1$ . We already know that value of $\cos \dfrac{\pi }{2}=0$; so after putting these values we get the following:
$\begin{align}
  & f\left( \dfrac{\pi }{4} \right)=\dfrac{{{\cot }^{3}}\dfrac{\pi }{4}-\tan \dfrac{\pi }{4}}{\cos \left( \dfrac{\pi }{2} \right)}\Rightarrow f\left( \dfrac{\pi }{4} \right)=\dfrac{\left( {{1}^{3}} \right)-\left( 1 \right)}{0} \\
 & \Rightarrow f\left( \dfrac{\pi }{4} \right)=\dfrac{0}{0} \\
\end{align}$
Since the limit is of $\dfrac{0}{0}$ form we will use the L’Hospital Rule
The L’Hospital Rule states that if $\phi (x)$ and $\psi (x)$takes the form$\dfrac{0}{0}$ then $\underset{x\to a}{\mathop{\lim }}\,\dfrac{\phi (x)}{\psi (x)}=\underset{x\to a}{\mathop{\lim }}\,\dfrac{\phi '(x)}{\psi '(x)}$
We have with us : \[\displaystyle \lim_{x \to \dfrac{\pi }{4}}\dfrac{{{\cot }^{3}}x-\tan x}{\cos \left( x+\dfrac{\pi }{4} \right)}\]
Applying the L’Hospital Rule into the limit below:
\[\begin{align}
  & \displaystyle \lim_{x \to \dfrac{\pi }{4}}\dfrac{{{\cot }^{3}}x-\tan x}{\cos \left( x+\dfrac{\pi }{4} \right)}\Rightarrow ~~\displaystyle \lim_{x \to \dfrac{\pi }{4}}\dfrac{\dfrac{d\left( {{\cot }^{3}}x \right)}{dx}-\dfrac{d\left( \tan x \right)}{dx}}{\dfrac{d\left( \cos \left( x+\dfrac{\pi }{4} \right) \right)}{dx}} \\
 & \Rightarrow \dfrac{3{{\cot }^{2}}x.\dfrac{d\left( \cot x \right)}{dx}-\dfrac{d\left( \tan x \right)}{dx}}{\dfrac{d\left( \cos \left( x+\dfrac{\pi }{4} \right) \right)}{dx}}.............\text{ Equation 4}\text{.} \\
 & \\
\end{align}\]
We will now find out the derivatives of cot x and tan x:
First we will find out the derivative of tan x:
\[\begin{align}
  & \dfrac{d\tan }{dx}=\dfrac{d\left( \dfrac{\sin x}{\cos x} \right)}{dx}\Rightarrow \dfrac{\left( \cos x.\dfrac{d\left( \sin x \right)}{dx} \right)-\left( \sin x.\dfrac{d\left( \cos x \right)}{dx} \right)}{{{\cos }^{2}}x} \\
 & \Rightarrow \dfrac{\left( \cos \left( x \right).\cos \left( x \right) \right)-\left( \sin \left( x \right).\left( -\sin \left( x \right) \right) \right)}{{{\cos }^{2}}x}=\dfrac{\left( \cos \left( x \right).\cos \left( x \right) \right)+\left( \sin \left( x \right).\sin \left( x \right) \right)}{{{\cos }^{2}}x} \\
 & =\dfrac{{{\cos }^{2}}\left( x \right)+{{\sin }^{2}}\left( x \right)}{{{\cos }^{2}}\left( x \right)}=\dfrac{1}{{{\cos }^{2}}x} \\
 & \therefore \dfrac{d\left( \tan x \right)}{dx}=\dfrac{1}{{{\cos }^{2}}x} \\
\end{align}\]

Now we will find out the derivative of cot x :
$\begin{align}
  & \dfrac{d\left( \cot x \right)}{dx}=\dfrac{d\left( \dfrac{1}{\tan x} \right)}{dx}\Rightarrow \dfrac{\left( \tan x.\dfrac{d\left( 1 \right)}{dx} \right)-\left( 1.\dfrac{d\left( \tan x \right)}{dx} \right)}{{{\tan }^{2}}x} \\
 & \Rightarrow \dfrac{0-{{\sec }^{2}}x}{{{\tan }^{2}}x}=\dfrac{-{{\sec }^{2}}x}{{{\tan }^{2}}x}=\dfrac{\left( \dfrac{-1}{{{\cos }^{2}}x} \right)}{\left( \dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x} \right)}=\dfrac{-1}{{{\sin }^{2}}x} \\
 & \therefore \dfrac{d\left( \cot x \right)}{dx}=\dfrac{-1}{{{\sin }^{2}}x} \\
\end{align}$

We already know that derivative of $\cos \theta =-\sin \theta $
Therefore, $\dfrac{d\left( \cos \left( x+\dfrac{\pi }{4} \right) \right)}{dx}=-\sin \left( x+\dfrac{\pi }{4} \right)$
Now we will put all the obtained derivatives in equation 4:
\[\begin{align}
  & \displaystyle \lim_{x \to \dfrac{\pi }{4}}\dfrac{{{\cot }^{3}}x-\tan x}{\cos \left( x+\dfrac{\pi }{4} \right)}\Rightarrow ~~\displaystyle \lim_{x \to \dfrac{\pi }{4}}\dfrac{\dfrac{d\left( {{\cot }^{3}}x \right)}{dx}-\dfrac{d\left( \tan x \right)}{dx}}{\dfrac{d\left( \cos \left( x+\dfrac{\pi }{4} \right) \right)}{dx}}\Rightarrow \dfrac{3{{\cot }^{2}}x.\dfrac{d\left( \cot x \right)}{dx}-\dfrac{d\left( \tan x \right)}{dx}}{\dfrac{d\left( \cos \left( x+\dfrac{\pi }{4} \right) \right)}{dx}} \\
 & \dfrac{3{{\cot }^{2}}x.\dfrac{d\left( \cot x \right)}{dx}-\dfrac{d\left( \tan x \right)}{dx}}{\dfrac{d\left( \cos \left( x+\dfrac{\pi }{4} \right) \right)}{dx}}=\dfrac{3{{\cot }^{2}}x.\left( \dfrac{-1}{{{\sin }^{2}}x} \right)-\dfrac{1}{{{\cos }^{2}}x}}{-\sin \left( x+\dfrac{\pi }{4} \right)} \\
 & \\
\end{align}\]
When we simplify the obtained term:\[\begin{align}
  & \displaystyle \lim_{x \to \dfrac{\pi }{4}}\dfrac{3{{\cot }^{2}}x.\left( \dfrac{-1}{{{\sin }^{2}}x} \right)-\dfrac{1}{{{\cos }^{2}}x}}{-\sin \left( x+\dfrac{\pi }{4} \right)}=\displaystyle \lim_{x \to \dfrac{\pi }{4}}\dfrac{3\dfrac{{{\cos }^{2}}x}{{{\sin }^{2}}x}.\left( \dfrac{-1}{{{\sin }^{2}}x} \right)-\dfrac{1}{{{\cos }^{2}}x}}{-\sin \left( x+\dfrac{\pi }{4} \right)} \\
 & \Rightarrow \displaystyle \lim_{x \to \dfrac{\pi }{4}}\dfrac{\dfrac{-3{{\cos }^{2}}x}{{{\sin }^{4}}x}-\dfrac{1}{{{\cos }^{2}}x}}{-\sin \left( x+\dfrac{\pi }{4} \right)}\Rightarrow \displaystyle \lim_{x \to \dfrac{\pi }{4}}\dfrac{\dfrac{-3{{\cos }^{4}}x-{{\sin }^{4}}x}{{{\sin }^{4}}x{{\cos }^{2}}x}}{-\sin \left( x+\dfrac{\pi }{4} \right)} \\
\end{align}\]
We now have: \[\displaystyle \lim_{x \to \dfrac{\pi }{4}}\dfrac{\dfrac{-3{{\cos }^{4}}x-{{\sin }^{4}}x}{{{\sin }^{4}}x{{\cos }^{2}}x}}{-\sin \left( x+\dfrac{\pi }{4} \right)}\] , As x is approaching $\dfrac{\pi }{4}$ as shown in the function, we will put as x as $\dfrac{\pi }{4}$ to find the limit:
\[\displaystyle \lim_{x \to \dfrac{\pi }{4}}\dfrac{\dfrac{-3{{\cos }^{4}}x-{{\sin }^{4}}x}{{{\sin }^{4}}x{{\cos }^{2}}x}}{-\sin \left( x+\dfrac{\pi }{4} \right)}=\dfrac{\dfrac{-3{{\cos }^{4}}\dfrac{\pi }{4}-{{\sin }^{4}}\dfrac{\pi }{4}}{{{\sin }^{4}}\dfrac{\pi }{4}{{\cos }^{2}}\dfrac{\pi }{4}}}{-\sin \left( \dfrac{\pi }{4}+\dfrac{\pi }{4} \right)}=\dfrac{\dfrac{-3{{\cos }^{4}}\dfrac{\pi }{4}-{{\sin }^{4}}\dfrac{\pi }{4}}{{{\sin }^{4}}\dfrac{\pi }{4}{{\cos }^{2}}\dfrac{\pi }{4}}}{-\sin \left( \dfrac{\pi }{2} \right)}\]
We know that \[\sin \left( \dfrac{\pi }{2} \right)=1\], Therefore \[\dfrac{\dfrac{-3{{\cos }^{4}}\dfrac{\pi }{4}-{{\sin }^{4}}\dfrac{\pi }{4}}{{{\sin }^{4}}\dfrac{\pi }{4}{{\cos }^{2}}\dfrac{\pi }{4}}}{-\sin \left( \dfrac{\pi }{2} \right)}=\dfrac{\dfrac{-3{{\cos }^{4}}\dfrac{\pi }{4}-{{\sin }^{4}}\dfrac{\pi }{4}}{{{\sin }^{4}}\dfrac{\pi }{4}{{\cos }^{2}}\dfrac{\pi }{4}}}{-1}\] ,
Now we know that \[\sin \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}\text{ and }\cos \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}\] , Putting these values above :
\[\begin{align}
  & \dfrac{-3{{\cos }^{4}}\dfrac{\pi }{4}-{{\sin }^{4}}\dfrac{\pi }{4}}{-1.{{\sin }^{4}}\dfrac{\pi }{4}{{\cos }^{2}}\dfrac{\pi }{4}}=\left( \dfrac{-3{{\left( \dfrac{1}{\sqrt{2}} \right)}^{4}}-{{\left( \dfrac{1}{\sqrt{2}} \right)}^{4}}}{-1.{{\left( \dfrac{1}{\sqrt{2}} \right)}^{4}}.{{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}} \right)=\left( \dfrac{-3\left( \dfrac{1}{4} \right)-\left( \dfrac{1}{4} \right)}{-\left( \dfrac{1}{4} \right).\left( \dfrac{1}{2} \right)} \right) \\
 & \Rightarrow \left( \dfrac{\left( \dfrac{-3-1}{4} \right)}{-\left( \dfrac{1}{8} \right)} \right)=\left( \dfrac{\left( \dfrac{-4}{4} \right)}{-\left( \dfrac{1}{8} \right)} \right)\Rightarrow \left( \dfrac{-1}{-1}\times 8 \right)\Rightarrow 8 \\
\end{align}\]
Therefore, \[\displaystyle \lim_{x \to \dfrac{\pi }{4}}\dfrac{{{\cot }^{3}}x-\tan x}{\cos \left( x+\dfrac{\pi }{4} \right)}=8\],

So, the correct answer is “Option C”.

Note: The derivative of tanx and cot x is found by first writing them in sin-cos form and then applying the Quotient Rule. Students should be conscious when deriving cot x and tan x, one can make mistakes for applying negative signs. There is no need to derive the derivative of tan x and cot x , students can directly use the standard values.
Before applying the L’Hospital Rule ensure that f(x) is of the form $\dfrac{0}{0};\dfrac{\infty }{\infty }$ then continue differentiation till you get the definite limit. Also keep in mind to differentiate the numerator and denominator separately.