0

Guest

Courses New

limit x tends to 0 then what is the value of 1-cosmx/1-cosnx=

limit x tends to 0 then what is the value of 1-cosmx/1-cosnx=

Grade:11

4 Answers

PASUPULETI GURU MAHESH
126 Points
8 years ago
limit x tends to 0   1-cosmx/1-cosnx= now rationalise nx))/(1-cos2nx)(1+cosnx)= ((1-cosmx)x1+cosnx/)x(1+cosnx(1-cosmx/1-cos
                                                          =(1+cosnx-cosmx.cosnx-cosmx)/(1-cos2nx)
                                                         =((1+cosnx).(1-cosmx))/(1-cos2nx)
                                                         =(1+cosnx)/(1-cos2nx)×(1-cosmx)
then solve that you get zerothere fore limit x tends to 0 then what is the value of 1-cosmx/1-cosnx=0
DEEPTHI JANGA
51 Points
8 years ago
(1-cosmx/1-cosnx)x(1+cosnx/1+cosnx)= ((1-cosmx)x(1+cosnx))/(1-cos2nx)1-cosmx/1-cosnx= now rationalise tends to 0   limit x
                                                          =(1+cosnx-cosmx.cosnx-cosmx)/(1-cos2nx)
                                                         =((1+cosnx).(1-cosmx))/(1-cos2nx)
                                                         =(1+cosnx)/(1-cos2nx)×(1-cosmx)
then solve that you get zerothere fore limit x tends to 0 then what is the value of 1-cosmx/1-cosnx=0 sorry but the options are a.mn  b.m/n   c.m2/n2    d.m-n       
SREEKANTH
85 Points
7 years ago
limit x tends to zero (1-cosmx)/(1-cosnx) by dividing the euation with mx and multiply with nx then you will get =limit x tends to zero (1-cosmx)/(1-cos nx)*(nx)/(mx) on re arranging we will get
              =((mx)(1-cosmx)/(mx)/(nx)(1-cos nx)/(nx)  but we know that 1-cos x=2sin^2 x/2
              =(nx/mx)(2(2sin^2 (mx)/2)/(mx)/2)/2(2sin^2 (nx)/2)/(nx)/2)
              =(n/m)(16/16)(n/m) limit x tends to zero ((sin (mx)/2)/(mx)/2))sin (mx)/2)/(mx)/2))/sin^2 (nx)/2)/(nx)/2)sin^2 (nx)/2)/(nx)/2)
              =(n/m)^2
 so the answer is (n/m)^2
Ajay
209 Points
7 years ago
Since this of form 0/0 use L Hospital rule. Differentiating numerator and denominator it becomes
 (m sin  mx)/ (n sin nx)
This is again of form 0/0 , hence differentiating again
 (m2 cos mx)/ (n2 cos nx)
as x tends to 0,  both cos mx and con nx tends to 1 hence m2/n2  should be correct answer

Think You Can Provide A Better Answer ?

Other Related Questions on Differential Calculus

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More