find limit of f(x)= (1/x+1)+1/x-1
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Answer:
No. Lateral limits are different at
x
0
=
0
lim
x
→
0
−
f
(
x
)
=
lim
x
→
0
−
e
1
x
+
1
e
1
x
−
1
=
−
1
lim
x
→
0
+
f
(
x
)
=
lim
x
→
0
+
e
1
x
+
1
e
1
x
−
1
=
1
so
lim
x
→
0
−
f
(
x
)
≠
lim
x
→
0
+
f
(
x
)
and
f
is as a result not continuous at
x
0
=
0
because
lim
x
→
0
−
e
1
x
+
1
e
1
x
−
1
1
x
=
y
x
→
0
−
y
→
−
∞
=
lim
y
→
−
∞
e
y
+
1
e
y
−
1
=
0
+
1
0
−
1
=
−
1
lim
x
→
0
+
e
1
x
+
1
e
1
x
−
1
1
x
=
u
x
→
0
+
u
→
+
∞
=
lim
u
→
+
∞
e
u
+
1
e
u
−
1
--
(
+
∞
+
∞
)
and with rules De L'Hopital
=
lim
u
→
+
∞
e
u
e
u
=
lim
u
→
+
∞
1
=
1
(Note: you can check in the graph the behaviour of
e
x
while
x
→
±
∞
)
graph{e^x [-22.79, 22.82, -11.42, 11.37]}
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