Question

1
X,
0 < x <
2.
1
2B. Let f(x) = { 0,
0, X
x =
show that Lt f(x) does not exist.
2.
x->-
2
x-1,
1
- < X < 1
2.​

Answer 1

Answer:

Given

x→0

lim

e

x

1

+1

e

x

1

−1

Left hand limit.

L.H.L =

x→0

−

lim

e

0−

1

+1

e

0−

1

−1

=

e

−∞

+1

e

−∞

−1

=

0+1

0−1

=−1

we know that e

∞

=∞

e

−∞

=

e

∞

1

=

∞

1

⇒0

Right hand limit

R.H.L =

x→0

+

lim

e

0+

1

+1

e

0+

1

−1

=

e

∞

+1

e

∞

−1

=

e

∞

(1+

e

∞

1

)

e

∞

(1−

e

∞

1

)

=

1+0

1−0

=1

So, here L.H.L



=R.H.L

So, limit does not exist