Continuity of a Function Example 1

Question 1

Discuss continuity of
$ \displaystyle f(x)=\left\{ \begin{array}{l}x+1,\text{ }x<1\\2,\text{ }\text{ }x<1\\{{x}^{2}},\text{ }x\ge 1\end{array} \right.$
at [x = 1].

Solution (step-by-step):

1. Compute the left-hand limit (LHL):
   [\lim_{x\to1^-} (x+1)] [=1+1=2.]
2. Compute the right-hand limit (RHL):
   [\lim_{x\to1^+} x^2] [=1^2=1.]
3. Value at the point: [f(1)=2].
4. Compare: LHL = 2, RHL = 1, and f(1) = 2. Since LHL ≠ RHL, the two-sided limit does not exist.
   Conclusion: [f] is discontinuous at [x=1] (jump discontinuity because left and right limits differ).

Question 2

Discuss continuity of
$ \displaystyle f(x)=\left\{ {\begin{array}{*{20}{c}} {\dfrac{{{{x}^{2}}-9}}{{x-3}},} & {x<3,} \\ {6,} & {x=3,} \\ {\dfrac{1}{{x-3}},} & {x>3,} \end{array}} \right.$
at [x = 3].

Solution (step-by-step):

1. Simplify the left piece for [x\ne3]:
   [\dfrac{x^2-9}{x-3}][=\dfrac{(x-3)(x+3)}{x-3}=x+3.]
   So for [x<3], [f(x)=x+3].
2. LHL:
   [\lim_{x\to3^-} (x+3)][=3+3=6.]
3. RHL:
   [
   \lim_{x\to3^+} \dfrac{1}{x-3}.
   ]
   As [x\to3^+] the denominator [+0] → the fraction → [+∞]. So RHL does not exist as a finite number (it is infinite).
4. f(3)=6. Compare: LHL = 6, RHL = ∞ (no finite limit), f(3)=6. Two-sided limit does not exist.
   Conclusion: [f] is discontinuous at [x=3] (infinite / non-finite behavior from right).

Question 3

Discuss continuity of
$ \displaystyle f(x)=\left\{ {\begin{array}{*{20}{c}} {\dfrac{{\sin x}}{x},} & {x<0,} \\ {1,} & {x=0,} \\ {\cos x,} & {x>0,} \end{array}} \right.$
at [x = 0].

Solution (step-by-step):

1. LHL: use standard limit
   [\lim_{x\to0^-} \dfrac{\sin x}{x}] [= 1.]
   (That standard limit is true from both sides.)
2. RHL:
   [\lim_{x\to0^+} \cos x][=\cos 0 = 1.]
3. f(0)=1.
4. Compare: LHL = 1, RHL = 1, f(0) = 1. All equal.
   Conclusion: [f] is continuous at [x=0] (removable-type pieces but defined consistently).

Question 4

Discuss continuity of
$ \displaystyle f(x)=\left\{ {\begin{array}{*{20}{c}} {\dfrac{{\mid x\mid }}{x},} & {x\ne 0,} \\ {0,} & {x=0,} \end{array}} \right.$
at [x = 0].

Solution (step-by-step):

1. For [x>0], [\dfrac{|x|}{x}=\dfrac{x}{x}=1]. For [x<0], [\dfrac{|x|}{x}=\dfrac{-x}{x}=-1].
2. RHL:
   [
   \lim_{x\to0^+}\dfrac{|x|}{x}=1.
   ]
3. LHL:
   [
   \lim_{x\to0^-}\dfrac{|x|}{x}=-1.
   ]
4. f(0)=0. Compare: LHL = -1, RHL = 1, f(0)=0. One-sided limits are not equal and neither equals f(0).
   Conclusion: [f] is discontinuous at [x=0] (jump discontinuity).

Question 5

Discuss continuity of
$ \displaystyle f(x)=\left\{ {\begin{array}{*{20}{c}} {{{x}^{2}},} & {x\le 2,} \\ {4x-4,} & {x>2,} \end{array}} \right.$
at [x = 2].

Solution (step-by-step):

1. LHL (and value from left):
   [\lim_{x\to2^-} x^2] [= 2^2 = 4,][\qquad] [f(2)=2^2=4.]
2. RHL:
   [\lim_{x\to2^+} (4x-4)][=4(2)-4][=8-4=4.]
3. Compare: LHL = 4, RHL = 4, f(2)=4. All equal.
   Conclusion: [f] is continuous at [x=2].