Answer:
The given function is f(x)=
⎩
⎪
⎨
⎧
x
sinx
,ifx<0
x+1,ifx≥0
It is evident that f is defined at all points of the real line.
Let c be a real number.
Case I
If c<0, then f(c)=
c
sinc
and
x→c
lim
f(x)=
=
Therefore, f is continuous at all points x, such that x<0
Case II
If c>0, then f(c)=c+1 and
(x+1)=c+1
∴
f(x)=f(c)
Therefore, f is continuous at all points such that x>0
Case III
If c=0, then f(c)=f(0)=0+1=1
The left hand limit of f at x=0 is,
x→0
=1
The right hand limit of f at x=0 is,
(x+1)=1
+
−
f(x)=f(0)
Therefore, f is continuous at x=0
From the above observations, it can be concluded that f is continuous at all points of the real line.
Thus, f has no point of discontinuity.
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Answers & Comments
Answer:
The given function is f(x)=
⎩
⎪
⎨
⎪
⎧
x
sinx
,ifx<0
x+1,ifx≥0
It is evident that f is defined at all points of the real line.
Let c be a real number.
Case I
If c<0, then f(c)=
c
sinc
and
x→c
lim
f(x)=
x→c
lim
x
sinx
=
c
sinc
Therefore, f is continuous at all points x, such that x<0
Case II
If c>0, then f(c)=c+1 and
x→c
lim
f(x)=
x→c
lim
(x+1)=c+1
∴
x→c
lim
f(x)=f(c)
Therefore, f is continuous at all points such that x>0
Case III
If c=0, then f(c)=f(0)=0+1=1
The left hand limit of f at x=0 is,
x→0
lim
f(x)=
x→0
lim
x
sinx
=1
The right hand limit of f at x=0 is,
x→0
lim
f(x)=
x→0
lim
(x+1)=1
∴
x→0
+
lim
f(x)=
x→0
−
lim
f(x)=f(0)
Therefore, f is continuous at x=0
From the above observations, it can be concluded that f is continuous at all points of the real line.
Thus, f has no point of discontinuity.