Ask a Doubt
Request a call back

Join NOW to get access to exclusive study material for best results

JEE Class main Answered

Find the limits of the following
question image
Asked by sarveshvibrantacademy | 05 May, 2019, 15:57: PM
answered-by-expert Expert Answer
L equals limit as x rightwards arrow 1 of open parentheses tan πx over 4 close parentheses to the power of tan πx over 2 end exponent space i s space o f space t h e space f o r m space 1 to the power of infinity rightwards double arrow L equals limit as x rightwards arrow 1 of open parentheses tan πx over 4 close parentheses to the power of tan πx over 2 end exponent space equals e to the power of limit as x rightwards arrow 1 of open parentheses tan πx over 2 close parentheses open parentheses tan πx over 4 minus 1 close parentheses end exponent N o w comma space limit as x rightwards arrow 1 of open parentheses tan πx over 2 close parentheses open parentheses tan πx over 4 minus 1 close parentheses equals limit as x rightwards arrow 1 of fraction numerator open parentheses tan πx over 4 minus 1 close parentheses over denominator cot πx over 2 end fraction T a k i n g space L apostrophe H o s p i t a l apostrophe s space r u l e comma space w e space h a v e limit as x rightwards arrow 1 of open parentheses tan πx over 2 close parentheses open parentheses tan πx over 4 minus 1 close parentheses equals limit as x rightwards arrow 1 of fraction numerator open parentheses sec squared πx over 4 cross times begin display style straight pi over 4 end style close parentheses over denominator negative cosec squared begin display style πx over 2 end style cross times begin display style straight pi over 2 end style end fraction equals limit as x rightwards arrow 1 of 1 half cross times fraction numerator sec squared open parentheses πx over 4 close parentheses over denominator negative cosec squared begin display style πx over 2 end style end fraction equals negative 1 half cross times open parentheses square root of 2 close parentheses squared over 1 squared equals negative 1 rightwards double arrow L equals e to the power of limit as x rightwards arrow 1 of open parentheses tan πx over 2 close parentheses open parentheses tan πx over 4 minus 1 close parentheses end exponent equals e to the power of negative 1 end exponent equals 1 over e H e n c e comma space limit as x rightwards arrow 1 of open parentheses tan πx over 4 close parentheses to the power of tan πx over 2 end exponent equals 1 over e
Answered by Renu Varma | 06 May, 2019, 11:48: AM

Application Videos

Concept Videos

Multiple Choice Questions Practice Test
Answered Unanswered My Questions My Answers
JEE main - Maths
in how many ways a team of 2 boys and 4 girls selected from 5 boys and 6 girls
Asked by pallavibirangal87 | 08 Dec, 2024, 10:18: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Maths
if a chord of length 18 cm is 12 cm away from the centre then find the distance of 24 cm long chord from the centre
Asked by pritisindra6 | 06 Dec, 2024, 21:04: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Maths
The number of ways in which at least one coin can be selected from 3 coins of one rupee, 4 coins of fifty paise, 5 coins of twenty paise and 6 coins of ten paise is-
Asked by kanishthkumawat | 29 Nov, 2024, 00:23: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Maths
what is range of tan^-1 x
Asked by pawandhayal125 | 06 Oct, 2024, 10:03: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Maths
what is 2+2
Asked by srishtipra2008 | 28 Sep, 2024, 10:33: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Maths
n arithmetic means are inserted between the numbers 7 and 49. if the sum of these means is 364 then the sum of their squares is
Asked by yogeshsoni575 | 23 Sep, 2024, 23:01: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Maths
consider a dodecagon then number of ways of selecting two triangles which can be made using the vertices of dodecagon so that they do not have a common side.
Asked by sumit.parsudih | 17 Sep, 2024, 22:22: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Maths
the sum of infinite series 1/2(1/3+1/4)-1/4(1/3^2+1/4^2)+1/6(1/3^3+1/4^3)+
Asked by pritamlovevanshi5 | 15 Sep, 2024, 08:24: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Maths
solve the equation 6x-y+3z=10 , 3x+6y=18 , x+y+5z=-10 by jacobis method with (3,0,-2) as the initial approcimate to the solution
Asked by ganeshbabugottapu7 | 12 Sep, 2024, 18:19: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Maths
find the inverse of a Matrix by using gauss Jordan method -2. 1. 3 A = 0. -1. 1 1. 2. 0
Asked by ganeshbabugottapu7 | 10 Sep, 2024, 07:32: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
My Class Videos Tests ₹ Plans Ask a Doubt