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Question: How do you determine the limit of \(\cot x\) as x approaches \({{\pi }^{-}}\)?...

How do you determine the limit of cot⁡x\cot x as x approaches π−{{\pi }^{-}}?

Explanation

Solution

To determine the limit of cot⁡x\cot x as x approaches π−{{\pi }^{-}}, we are going to first write cot⁡x=cos⁡xsin⁡x\cot x=\dfrac{\cos x}{\sin x} and then as x approaches to π−{{\pi }^{-}} so we are going to take “h” which lies between 0 and 1 and this value of h is very less than 1. Then we write in place of (x−h)\left( x-h \right) in place of x in cos⁡xsin⁡x\dfrac{\cos x}{\sin x} and then put the limit h tending to 0. And then simplify.

Complete step-by-step answer:
In the above problem, it is given that cot⁡x\cot x as x approaches π−{{\pi }^{-}} so writing this limit expression in the mathematical form we get,
⇒lim⁡x→π−cot⁡x\Rightarrow \displaystyle \lim_{x\to {{\pi }^{-}}}\cot x
Now, we know from the trigonometric properties that:
cot⁡x=cos⁡xsin⁡x\cot x=\dfrac{\cos x}{\sin x}
So, using the above relation in the above limit we get,
⇒lim⁡x→π−cos⁡xsin⁡x\Rightarrow \displaystyle \lim_{x\to {{\pi }^{-}}}\dfrac{\cos x}{\sin x}
It is given that x approaches π−{{\pi }^{-}} so this means that x value is a subtraction of π\pi with some number. Let us assume that a small number is “h”. This number “h” lies between 0 and 1 in the following way:
0<h<<10 < h << 1
So, we can write x=π−hx=\pi -h in the above limit where h is approaching 0.
⇒lim⁡h→0cos⁡(π−h)sin⁡(π−h)\Rightarrow \displaystyle \lim_{h \to 0}\dfrac{\cos \left( \pi -h \right)}{\sin \left( \pi -h \right)}
We know the trigonometric identities of sine and cosine as follows:
cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B; sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B \begin{aligned} & \cos \left( A-B \right)=\cos A\cos B+\sin A\sin B; \\\ & \sin \left( A-B \right)=\sin A\cos B-\cos A\sin B \\\ \end{aligned}
Using the above trigonometric identities in the above limit we get,
⇒lim⁡h→0cos⁡πcosh⁡+sin⁡πsinh⁡sin⁡πcosh⁡−cos⁡πsinh⁡\Rightarrow \displaystyle \lim_{h \to 0}\dfrac{\cos \pi \cosh +\sin \pi \sinh }{\sin \pi \cosh -\cos \pi \sinh }
We know the values of sine of π\pi and cosine of π\pi as follows:
sin⁡π=0; cos⁡π=−1 \begin{aligned} & \sin \pi =0; \\\ & \cos \pi =-1 \\\ \end{aligned}
Substituting the above values in the limit expression we get,
⇒lim⁡h→0(−1)cosh⁡+(0)sinh⁡(0)cosh⁡−(−1)sinh⁡ ⇒lim⁡h→0−cosh⁡sinh⁡ \begin{aligned} & \Rightarrow \displaystyle \lim_{h \to 0}\dfrac{\left( -1 \right)\cosh +\left( 0 \right)\sinh }{\left( 0 \right)\cosh -\left( -1 \right)\sinh } \\\ & \Rightarrow \displaystyle \lim_{h \to 0}\dfrac{-\cosh }{\sinh } \\\ \end{aligned}
Applying the value of limit by putting h as 0 in the above fraction we get,
−cos⁡0sin⁡0-\dfrac{\cos 0}{\sin 0}
We know that the value of sin⁡0=0&cos⁡0=1\sin 0=0\And \cos 0=1 so substituting these values in the above w eget,
−10-\dfrac{1}{0}
And we know that 10\dfrac{1}{0} is not defined or infinity.
−∞-\infty
Hence, the evaluation of the above limit is −∞-\infty .

Note: The possible mistake in the above problem is that in the last two steps you might forget to put a negative sign in front of the infinity because you might think what difference does it make if we remove this negative sign in front of the infinity.
−10 =∞ \begin{aligned} & -\dfrac{1}{0} \\\ & =\infty \\\ \end{aligned}
This is the wrong answer because positive and negative infinities are completely different. As one is pointing in a positive direction and the other is in a negative direction so make sure you won’t make this mistake.