Question

1-cos x /2x^2 find the value at limit x=0​

Answer 1

Answer:

L'Hopital's rules says that the limx→af(x)g(x)⇒f'(a)g'(a)

Using this, we get limx→01−cosxx2⇒−sin02(0)

Yet as the denominator is 0, this is impossible. So we do a second limit:

lim(x→0)sinx2x⇒cos02=12=0.5

So, in total

limx→01−cosxx2⇒limx→0sinx2x⇒cosx2⇒cos02=12

Step-by-step explanation:

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