Limit x _1. 1+ lnx-x/1-2x+ x2
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Answer:
If
lim
x
→
c
f
(
x
)
=
0
and
lim
x
→
c
g
(
x
)
=
0
and
lim
x
→
c
f
'
(
x
)
g
'
(
x
)
=
L
Then
lim
x
→
c
f
(
x
)
g
(
x
)
=
L
In our case,
f
(
x
)
=
ln
(
x
)
and
g
(
x
)
=
x
−
1
. Both their derivatives are rather straight-forward:
f
'
(
x
)
=
1
x
g
'
(
x
)
=
1
This means we can rewrite our limit like so:
lim
x
→
1
ln
(
x
1
=
1
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