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Question: How do you find the limit of \(arctan\left( x \right)\) as \(x\) approaches infinity?...

How do you find the limit of arctan(x)arctan\left( x \right) as xx approaches infinity?

Explanation

Solution

The question is solved by finding out the principal value of the function by putting x=∞x=\infty . Then the general limit of the inverse tan function is obtained which is the sum of principal value and nπn\pi . As the values of x are tending to positive infinity only the specific range of the general limit is considered.

Complete step by step solution:
The arctan⁡(x)\arctan \left( x \right) is the inverse of the tan⁡x\tan x function. It returns an angle for the numerical value of the tangent given.
⇒Arctan(x) = 1tan(x)\Rightarrow Arctan\left( x \right)\text{ }=\text{ }\dfrac{1}{tan\left( x \right)}
The domain of the trigonometric function is the set of all real numbers and the function ranges from −π2Thegraphoftheinversetangentfunctionissymmetricalaboutthex−axisandisboundedbetween-\dfrac{\pi }{2}The graph of the inverse tangent function is symmetrical about the x-axis and is bounded between -\dfrac{\pi }{2}andand\dfrac{\pi }{2}Thegraphneitherhasanymaximaorminimahavingaperiodof The graph neither has any maxima or minima having a period of\piThegeneralsolutionorprincipalvaluesoftheinversetangentfunctionisgivenas, The general solution or principal values of the inverse tangent function is given as, \Rightarrow 2\pi k~+~\dfrac{\pi }{2};as; askapproachespositiveinfinity.Theaboveequationcanbesimplifiedas,approaches positive infinity. The above equation can be simplified as, \Rightarrow \pi \left( 2k\text{ }+\text{ }\dfrac{1}{2} \right)Letusassumethat Let us assume thaty\text{ }=\text{ }arctan\left( x \right)Accordingtothequestion,Fromthetrigonometrictable,weknowthat According to the question, From the trigonometric table, we know thatarctan\left( \infty \right)isis\dfrac{\pi }{2}Writingthesameweget, Writing the same we get, \displaystyle \lim_{x \to \dfrac{\pi }{2}}tanx=+\inftyNowtakingthetangentontheotherside, Now taking the tangent on the other side, \Rightarrow \displaystyle \lim_{x \to +\infty }y=\dfrac{\pi }{2}Asthevalueofxtendstoinfinitytheprincipalvalueofyobtainedis As the value of x tends to infinity the principal value of y obtained is\dfrac{\pi }{2}Thegenerallimitfortheabovetrigonometricfunctionisgivenasthesumofprincipalvalueandkπwhere The general limit for the above trigonometric function is given as the sum of principal value and kπ wherek=0,\pm 1,\pm 2.....Hencethevaluesofxapproachingpositiveinfinitythevaluesofkmustbeevenwhichmeansthegenerallimitisequalto Hence the values of x approaching positive infinity the values of k must be even which means the general limit is equal to\dfrac{\pi }{2}\text{ }+\text{ }2k\piwherewherek=0,\pm 1,\pm 2.....$.

Note: We must know the formula for the general solutions of arctan(x)arctan\left( x \right) to solve the question in an easy way. We must note that the general limit of the function is given by the principal value +nπ+n\pi. The general limit of the values of x approaching negative infinity is given by π2 + (2k+1)π2\dfrac{\pi }{2}\text{ }+\text{ }\left( 2k+1 \right)\dfrac{\pi }{2} where k=0,±1,±2.....k=0,\pm 1,\pm 2..... that is only the odd integral of π\pi.