Value of limit x tends to 0 (1-cos2x)^1/2/x
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Hi Mate!!!
√{ ( 1 - Cos ( 2x ) } = 2 Sin ( x )
=>. Lt 2 Sin ( x ) / x = 2
x--->0
And
=>. Lt - 2 Sin ( x ) / x = -2
x--->0
So, limit doesn't exist
√{ ( 1 - Cos ( 2x ) } = 2 Sin ( x )
=>. Lt 2 Sin ( x ) / x = 2
x--->0
And
=>. Lt - 2 Sin ( x ) / x = -2
x--->0
So, limit doesn't exist
Anonymous:
kya
wt?
kyu? ka kaha usna?
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