We have, f (x) = {

2x−3	x≤0
3(x+1)	x>0

For

lim

x→0

f (x) , we have to evaluate

lim

x→0−

f (x) and

lim

x→0+

f (x) separately as f(x) is defined differently on left and right of 0.
Now,

lim

x→0−

f (x) =

lim

x→0−

(2x + 3) = 2 . (0) + 3 = 3
and

lim

x→0+

f (x) =

lim

x→0+

3 (x + 1) = 3 (0 + 1) = 3.
Thus we see,

lim

x→0−

f (x) =

lim

x→0+

f (x) = 3
∴

lim

x→0

f (x) = 3
Now,

lim

x→1

f (x) 3 (x + 1) = 3 × 2 = 6
∴

lim

x→1

f (x) = 6 and

lim

x→0

f (x) = 3.