1It can b written as:
lim x→0 sin xn . (x)m . xn /(sin x)m.(x)m.xn
lim x→0 sin xn.(x)m.xn-m/xn.(sin x)m
Applying limits...
=0n-m = 0 Ans...
5 Answers
1
11I don't think there's any need of multiplying and dividing by xm....
f(x)=xnsinxm=0, as x goes to 0....
1as x tends to 0 sinx approaches x
as x tends to 0 sinxn approaches xn
as x tends to 0 (sinx)m approaches (x)n
lim x→0 xn / xm = xn-m
it is in the form of 0+ which tends to 0
had it been in form of 0-ve = 1/0+ve it wud have tended to infinity
1Hey,champ.u know Lt.sinx/x = 1. thats what anchal has done.
x→0
as,there is sin xn and sinmx, so multiply and divide xn in the first and xm,in the other.thus u get:
lt. xn-m = 0. and a better objective method has been given by Dubey sir
x→0