Question

find limit of f(x)= (1/x+1)+1/x-1

Answer 1

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]}