We have f (x) = {

mx2+n	x<0
nx+m	0≤x≤1
nx3+m	x>1

Now

lim

x→0−

f (x) =

lim

x→0−

(mx2+n) = m (0) + n = n
and

lim

x→0+

f (x) =

lim

x→0+

(nx + m) = n (0) + m = m
But for

lim

x→0

f (x) to exist, we must have

lim

x→0−

f (x) =

lim

x→0+

f (x)
i.e., n = m
Hence

lim

x→0

f (x) exists only if n = m.
Now,

lim

x→1−

f (x) =

lim

x→1−

(nx + m) = n + m
and

lim

x→1+

f (x) =

lim

x→1+

(nx3+m) = n + m
∴ Above condition shows that

lim

x→1−

f (x) =

lim

x→1+

f (x) = m + n. Thus

lim

x→1

f (x) exists for all n , m.