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Question: Calculate the following limit: \[\displaystyle \lim_{x \to \dfrac{\pi }{4}}\dfrac{{{\cot }^{3}}x-\ta...

Calculate the following limit: lim⁡x→π4cot⁡3x−tan⁡xcos⁡(x+π4)\displaystyle \lim_{x \to \dfrac{\pi }{4}}\dfrac{{{\cot }^{3}}x-\tan x}{\cos \left( x+\dfrac{\pi }{4} \right)}
A). 44
B). 828\sqrt{2}
C). 88
D). 424\sqrt{2}

Explanation

Solution

We will first find out whether the given function is in the form of (00)\left( \dfrac{0}{0} \right) or (∞∞)\left( \dfrac{\infty }{\infty } \right) by putting x as π4\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 π4\dfrac{\pi }{4} in the function to get the final limit

Complete step-by-step answer:
To find the limit of the given function: lim⁡x→π4cot⁡3x−tan⁡xcos⁡(x+π4)\displaystyle \lim_{x \to \dfrac{\pi }{4}}\dfrac{{{\cot }^{3}}x-\tan x}{\cos \left( x+\dfrac{\pi }{4} \right)}
As x is approaching π4\dfrac{\pi }{4} as shown in the function, we will put as x as π4\dfrac{\pi }{4} and see the value that comes:
Let’s take f(x)=cot⁡3x−tan⁡xcos⁡(x+π4)f\left( x \right)=\dfrac{{{\cot }^{3}}x-\tan x}{\cos \left( x+\dfrac{\pi }{4} \right)}, now when we put x as
f(π4)=cot⁡3π4−tan⁡π4cos⁡(π4+π4)⇒f(π4)=cot⁡3π4−tan⁡π4cos⁡(π2)............ Equation 1.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:
cot⁡θ=cos⁡θsin⁡θ⇒cot⁡π4=cos⁡(π4)sin⁡(π4) ...........Equation 2.  \begin{aligned} & \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{aligned}
We already know that the value of cos⁡π4=12\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}} and sin⁡π4=12\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}} , so we will putting these values in Equation 2 in order to find value of cot⁡π4\cot \dfrac{\pi }{4}:
cot⁡π4=cos⁡(π4)sin⁡(π4)⇒cot⁡π4=1212\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 12\dfrac{1}{\sqrt{2}} , we get cot⁡π4=1\cot \dfrac{\pi }{4}=1.
Similarly we will find out the value of tan⁡π4\tan \dfrac{\pi }{4} :
tan⁡θ=sin⁡θcos⁡θ⇒tan⁡π4=sin⁡(π4)cos⁡(π4) ...........Equation 3.\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⁡π4=12\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}} and sin⁡π4=12\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⁡π4\tan \dfrac{\pi }{4}:
tan⁡π4=sin⁡(π4)cos⁡(π4)⇒tan⁡π4=1212\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 12\dfrac{1}{\sqrt{2}} , we get tan⁡π4=1\tan \dfrac{\pi }{4}=1.
Putting the values of tan⁡π4\tan \dfrac{\pi }{4} and cot⁡π4\cot \dfrac{\pi }{4} in Equation 1.
We have: f(π4)=cot⁡3π4−tan⁡π4cos⁡(π2)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⁡π4=1\cot \dfrac{\pi }{4}=1 and tan⁡π4=1\tan \dfrac{\pi }{4}=1 . We already know that value of cos⁡π2=0\cos \dfrac{\pi }{2}=0; so after putting these values we get the following:
f(π4)=cot⁡3π4−tan⁡π4cos⁡(π2)⇒f(π4)=(13)−(1)0 ⇒f(π4)=00 \begin{aligned} & 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{aligned}
Since the limit is of 00\dfrac{0}{0} form we will use the L’Hospital Rule
The L’Hospital Rule states that if ϕ(x)\phi (x) and ψ(x)\psi (x)takes the form00\dfrac{0}{0} then lim⁡x→a ϕ(x)ψ(x)=lim⁡x→a ϕ′(x)ψ′(x)\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 : lim⁡x→π4cot⁡3x−tan⁡xcos⁡(x+π4)\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:

& \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{aligned}$$ We will now find out the derivatives of cot x and tan x: First we will find out the derivative of tan x: $$\begin{aligned} & \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{aligned}$$ Now we will find out the derivative of cot x : $\begin{aligned} & \dfrac{d\left( \cot x \right)}{dx}=\dfrac{d\left( \dfrac{1}{\tan x} \right)}{dx}\Rightarrow \dfrac{\left( \tan x.\dfrac{d\left( 1 \right)}{dx} \right)-\left( 1.\dfrac{d\left( \tan x \right)}{dx} \right)}{{{\tan }^{2}}x} \\\ & \Rightarrow \dfrac{0-{{\sec }^{2}}x}{{{\tan }^{2}}x}=\dfrac{-{{\sec }^{2}}x}{{{\tan }^{2}}x}=\dfrac{\left( \dfrac{-1}{{{\cos }^{2}}x} \right)}{\left( \dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x} \right)}=\dfrac{-1}{{{\sin }^{2}}x} \\\ & \therefore \dfrac{d\left( \cot x \right)}{dx}=\dfrac{-1}{{{\sin }^{2}}x} \\\ \end{aligned}$ We already know that derivative of $\cos \theta =-\sin \theta $ Therefore, $\dfrac{d\left( \cos \left( x+\dfrac{\pi }{4} \right) \right)}{dx}=-\sin \left( x+\dfrac{\pi }{4} \right)$ Now we will put all the obtained derivatives in equation 4: $$\begin{aligned} & \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{aligned}$$ When we simplify the obtained term:$$\begin{aligned} & \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{aligned}$$ We now have: $$\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{aligned} & \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{aligned}$$ Therefore, $$\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.