{"id":3958,"date":"2021-08-11T12:52:23","date_gmt":"2021-08-11T12:52:23","guid":{"rendered":"https:\/\/mathemerize.com\/?p=3958"},"modified":"2021-11-26T16:37:56","modified_gmt":"2021-11-26T11:07:56","slug":"limit-of-trigonometric-functions","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/limit-of-trigonometric-functions\/","title":{"rendered":"How to Find Limit of Trigonometric Functions"},"content":{"rendered":"
Here, you will learn how to find limit of trigonometric functions and limits using series expansion with example.<\/p>\n
Let’s begin –\u00a0<\/p>\n
\\(\\displaystyle{\\lim_{x \\to 0}}\\) \\(sinx\\over x\\) = 1 = \\(\\displaystyle{\\lim_{x \\to 0}}\\) \\(tanx\\over x\\) = \\(\\displaystyle{\\lim_{x \\to 0}}\\) \\(tan^{-1}x\\over x\\) = \\(\\displaystyle{\\lim_{x \\to 0}}\\) \\(sin^{-1}x\\over x\\) [where x is measured in radians]<\/p>\n
(a)\u00a0 If \\(\\displaystyle{\\lim_{x \\to a}}\\) f(x) = 0, then \\(\\displaystyle{\\lim_{x \\to a}}\\) \\(sinf(x)\\over f(x)\\) = 1
e.g. \\(\\displaystyle{\\lim_{x \\to 1}}\\) \\(sin(lnx)\\over (lnx)\\) = 1\u00a0<\/p>\n\n\n
Example : <\/span>Evaluate : \\(\\displaystyle{\\lim_{x \\to 0}}\\) \\(x^3 cotx\\over {1-cosx}\\)<\/p>\n Solution : <\/span>\n \\(\\displaystyle{\\lim_{x \\to 0}}\\) \\(x^3 cosx\\over {sinx(1-cosx)}\\) = \n\t\t \\(\\displaystyle{\\lim_{x \\to 0}}\\) \\(x^3 cosx(1 + cosx)\\over {sinxsin^2x}\\) = \n\t\t \\(\\displaystyle{\\lim_{x \\to 0}}\\) \\({x^3\\over sin^3x}.cosx(1 + cosx)\\) = 2 Example : <\/span> Evaluate : \\(\\displaystyle{\\lim_{x \\to 0}}\\) \\((2+x)sin(2+x)-2sin2\\over x\\)<\/p>\n Solution : <\/span>\\(\\displaystyle{\\lim_{x \\to 0}}\\) \\(2(sin(2+x)-sin2)+xsin(2+x)\\over x\\) Example : <\/span>Evaluate : \\(\\displaystyle{\\lim_{x \\to 0}}\\) \\(xln(1+2tanx)\\over 1-cosx\\)<\/p>\n Solution : <\/span>\\(\\displaystyle{\\lim_{x \\to 0}}\\) \\(xln(1+2tanx)\\over 1-cosx\\) Expansion of function like binomial expansion, exponential & logarithmic expansion, expansion of sinx, cosx, tanx should be remembered by heart which are given below :<\/p>\n (a) \\(e^x\\) = 1 + \\(x\\over 1!\\) + \\(x^2\\over {2!}\\) + ……..<\/p>\n (b) ln(1 + x) = x – \\(x^2\\over 2\\) + \\(x^3\\over 3\\) – \\(x^4\\over 4\\) + ………for -1 < x \\(\\leq\\) 1<\/p>\n (c) sinx = x – \\(x^3\\over 3!\\) + \\(x^5\\over 5!\\) – \\(x^7\\over 7!\\) + ……….<\/p>\n (d) cosx = 1 – \\(x^2\\over 2!\\) + \\(x^4\\over 4!\\) + \\(x^6\\over 6!\\) + ……….<\/p>\n (e) tanx = x + \\(x^3\\over 3\\) + \\(2x^5\\over 15\\) + …….<\/p>\n Hope you learnt how to find the limit of trigonometric functions. To learn more practice more questions and get ahead in competition. Good Luck!<\/p>\n\n\n
<\/p>\n\n\n\n
\n = \\(\\displaystyle{\\lim_{x \\to 0}}\\)(\\(2.2.cos(2+{x\\over 2})sin{x\\over 2}\\over x\\) + sin(2+x))
= \\(\\displaystyle{\\lim_{x \\to 0}}\\)\\(2cos(2+{x\\over 2})sin{x\\over 2}\\over {x\\over 2}\\) + \\(\\displaystyle{\\lim_{x \\to 0}}\\)sin(2+x)
= 2cos2 + sin2
<\/p>\n\n\n\n
= \\(\\displaystyle{\\lim_{x \\to 0}}\\) \\(xln(1+2tanx)\\over {1-cosx\\over x^2}.x^2\\).\\(2tanx\\over 2tanx\\)
= 4
<\/p>\n\n\nLimit using series expansion<\/h2>\n