lim (x²-1) (2. Cosx)
x_0
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Answer: L'Hopital's rules says that the
lim
x
→
a
f
(
x
)
g
(
x
)
⇒
f
'
(
a
)
g
'
(
a
)
Using this, we get
lim
x
→
0
1
−
cos
x
x
2
⇒
−
sin
0
2
(
0
)
Yet as the denominator is
0
, this is impossible. So we do a second limit:
lim
(
x
→
0
)
sin
x
2
x
⇒
cos
0
2
=
1
2
=
0.5
So, in total
lim
x
→
0
1
−
cos
x
x
2
⇒
lim
x
→
0
sin
x
2
x
⇒
cos
x
2
⇒
cos
0
2
=
1
2
Step-by-step explanation: PLEASE MARK ME THE BRAINLEIST
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