Additional Questions [BCB, page 394]

1. Define the continuity of a function at a point. Give with reason, an example of a continuouss function at a point. Is the function \(f(x)=\frac{1}{1-x}\) continuous at the point x=1?

   Solution 👉 Click Here

   Definition

   Let 𝑓(𝑥) be a function defined at all values in an open interval containing a
   Then we say that f(x) is continuous at a if
   for a given number \(\epsilon > 0\), there exists a number \(\delta > 0\)such that
   |f(x)-f(a)|<\(\epsilon\) whenever |x-a|<\(\delta\)

   OR
   

   the function f(x) is continuous at a if the following three conditions holds

   1. f(a) is defined
   2. \( \displaystyle \lim_{x\to a^{-}}f(x)\) exist = LHL
   3. \( \displaystyle \lim_{x\to a^{+}}f(x) \) exist = RHL
   4. LHL = RHL = f(a)

   Example with reason

   Let \( f(x)= \begin{cases} -x^2 +2 &\text { for } x \ne 2 \\ -2 &\text { for } x=2 \end{cases} \) thn f(x) is continuous at x=2.

   Because
   At \( x=2^{-}\), the left hand limit is
   \( \displaystyle \lim_{x \to 2^{-}} f(x)= \lim_{x \to 2^{-} } -x^2+2 =-2\) LHL
   At \( x=2^{+}\), the right hand limit is
   \( \displaystyle \lim_{x \to 2^{+}} f(x)= \lim_{x \to 2^{+}} = -x^2+2 =-2\) RHL
   At x=2, the value of the function is
   f(2) = -2 Functional Value
   Since, limiting value and functional value are equal, f(x) is continuous at x=2.

2. When a function f(x) is said to be continuous at a point x=a? Discuss the continuity of \(f(x)=\begin{cases} x^2+2 & \text {for } x \le 5 \\ 3x+12 & \text{for } x > 5 \end{cases}\) at x=5

   Solution 👉 Click Here

   Continuous

   Let 𝑓(𝑥) be a function defined at all values in an open interval containing a
   Then we say that f(x) is continuous at a if
   for a given number \(\epsilon > 0\), there exists a number \(\delta > 0\)such that
   |f(x)-f(a)|<\(\epsilon\) whenever |x-a|<\(\delta\)

   OR
   

   the function f(x) is continuous at a if the following three conditions holds

   1. f(a) is defined
   2. \( \displaystyle \lim_{x\to a^{-}}f(x)\) exist = LHL
   3. \( \displaystyle \lim_{x\to a^{+}}f(x) \) exist = RHL
   4. LHL = RHL = f(a)

   Problem Solving

   Given \(f(x)=\begin{cases} x^2+2 & \text {for } x \le 5 \\ 3x+12 & \text{for } x > 5 \end{cases}\) at x=5.

   Here
   At \( x=5^{-}\), the left hand limit is
   \( \displaystyle \lim_{x \to 5^{-}} f(x)= \lim_{x \to 5^{-} } x^2+2 =27\) LHL
   At \( x=5^{+}\), the right hand limit is
   \( \displaystyle \lim_{x \to 5^{+}} f(x)= \lim_{x \to 5^{+}} = 3x+12 =27\) RHL
   At x=5, the value of the function is
   \(f(2) =x^2+2= 27\) Functional Value
   Since, limiting value and functional value are equal, f(x) is continuous at x=5.

3. At what point is the function \(f(x)=\frac{x+1}{(x-2)(x-3)}\) (i) discontinuous (ii) continuous ?

   Solution 👉 Click Here

   Solution
   The function \(f(x)=\frac{x+1}{(x-2)(x-3)}\) is (i) discontinuous at x=2 and x=3 (ii) is continuous in other points
4. Doiscuss the continuity of the function f(x) at the point x=0.
   \(f(x)=\begin{cases} x & \text {for } x > 0 \\ 1 & \text{for } x=0 \\ -x & \text{for } x < 0 \end{cases}\)

   Solution 👉 Click Here

   Given \(f(x)=\begin{cases} x & \text {for } x > 0 \\ 1 & \text{for } x=0 \\ -x & \text{for } x < 0 \end{cases}\)

   Here
   At \( x=0^{-}\), the left hand limit is
   \( \displaystyle \lim_{x \to 0^{-}} f(x)= \lim_{x \to 0^{-} } -x =0\) LHL
   At \( x=0^{+}\), the right hand limit is
   \( \displaystyle \lim_{x \to 0^{+}} f(x)= \lim_{x \to 0^{+}} = x =0\) RHL
   At x=0, the value of the function is
   \(f(0) =1\) Functional Value
   Since, limiting value and functional value are NOT equal, f(x) is NOT continuous at x=0.

5. A function f(x) is defind as follows. \(f(x)=\begin{cases} 2x+1 & \text {for } x < 1 \\ 2 & \text{for } x=1 \\ 3x & \text{for } x > 0 \end{cases}\)
   Calculate the left hand limit and right hand limit of f(x) at x=1. Is the function continuous at x=1?

   Solution 👉 Click Here

   Given \(f(x)=\begin{cases} 2x+1 & \text {for } x < 1 \\ 2 & \text{for } x=1 \\ 3x & \text{for } x > 0 \end{cases}\)

   Here
   At \( x=1^{-}\), the left hand limit is
   LHL=\( \displaystyle \lim_{x \to 1^{-}} f(x)= \lim_{x \to 1^{-} } 2x+1 =\) 3
   At \( x=1^{+}\), the right hand limit is
   RHL=\( \displaystyle \lim_{x \to 1^{+}} f(x)= \lim_{x \to 1^{+}} = 3x =\) 3
   At x=1, the value of the function is
   F=\(f(1) =\) 2
   Since, limiting value and functional value are NOT equal, f(x) is NOT continuous at x=1.

6. What do you understand by the limit of a function? Let a function f(x) is defined by
   \(f(x)=\begin{cases} 2-x^2 & \text {for } x < 2 \\ 3 & \text{for } x=2 \\ x-4 & \text{for } x > 2 \end{cases}\)
   Verify that the limit of the function f(x) exists at x=2. Is the function f(x) continuous at x=2? If not why? State how can you make it continuous.

   Solution 👉 Click Here

   Given \(f(x)=\begin{cases} 2-x^2 & \text {for } x < 2 \\ 3 & \text{for } x=2 \\ x-4 & \text{for } x > 2 \end{cases}\)

   Here
   At \( x=2^{-}\), the left hand limit is
   LHL=\( \displaystyle \lim_{x \to 2^{-}} f(x)= \lim_{x \to 2^{-} } 2-x^2 =\) -2
   At \( x=1^{+}\), the right hand limit is
   RHL=\( \displaystyle \lim_{x \to 2^{+}} f(x)= \lim_{x \to 2^{+}} = x-4=\) -2
   Since,LHL=RHL, the limit of the function f(x) exists at x=2
   Next,
   At x=2, the value of the function is
   F=\(f(2) =\) 3
   The function f(x) is NOT continuous at x=2. Because, limiting value and functional value are NOT equal.
   We can make the function f(x) continuous by re-defining it as below.
   \(f(x)=\begin{cases} 2-x^2 & \text {for } x < 2 \\ -2 & \text{for } x=2 \\ x-4 & \text{for } x > 2 \end{cases}\)

7. A function f(x) is defined as under \(f(x)=\begin{cases} \frac{x^2-x-6}{x^2-2x-3} & \text {for } x \ne 3 \\ \frac{5}{3} & \text{for } x =3 \end{cases}\)
   Prove that f(x) is discontinuous at x=3. Can the definition of f(x) for x=3 be modified so as to make it continuous there?

   Solution 👉 Click Here

   Given \(f(x)=\begin{cases} \frac{x^2-x-6}{x^2-2x-3} & \text {for } x \ne 3 \\ \frac{5}{3} & \text{for } x =3 \end{cases}\)

   Here
   At \( x=3^{-}\), the left hand limit is
   LHL=\( \displaystyle \lim_{x \to 3^{-}} f(x)= \lim_{x \to 3^{-} } \frac{x^2-x-6}{x^2-2x-3}=\lim_{x \to 3^{-} }\frac{(x-3)(x+2)}{(x-3)(x+1)} =\lim_{x \to 3^{-} }\frac{x+2}{x+1}=\) \(\frac{5}{4}\)
   At \( x=3^{+}\), the right hand limit is
   RHL=\( \displaystyle \lim_{x \to 3^{+}} f(x)= \lim_{x \to 3^{+}} \frac{x^2-x-6}{x^2-2x-3}=\lim_{x \to 3^{+} }\frac{(x-3)(x+2)}{(x-3)(x+1)} =\lim_{x \to 3^{+} }\frac{x+2}{x+1}=\) \(\frac{5}{4}\)
   At x=3, the value of the function is
   F=\(f(3) =\) \(\frac{5}{3}\)
   The function f(x) is discontinuous at x=3. Because, limiting value and functional value are NOT equal.
   We can make the function f(x) continuous by re-defining it as below.
   \(f(x)=\begin{cases} \frac{x^2-x-6}{x^2-2x-3} & \text {for } x \ne 3 \\ \frac{5}{4} & \text{for } x =3 \end{cases}\)

8. 
   
   1. A function f(x) is defined as follows
      \(f(x)=\begin{cases} \frac{1}{2}+x & \text {for } 0 < x <\frac{1}{2} \\ \frac{1}{2} & \text{for } x =\frac{1}{2} \\ \frac{3}{2}-x & \text{for } \frac{1}{2} < x < 1 \end{cases}\)
      Show that f(x) has removable discontinuity at \(x=\frac{1}{2}\)

      Solution 👉 Click Here

      Given \(f(x)=\begin{cases} \frac{1}{2}+x & \text {for } 0 < x <\frac{1}{2} \\ \frac{1}{2} & \text{for } x =\frac{1}{2} \\ \frac{3}{2}-x & \text{for } \frac{1}{2} < x < 1 \end{cases}\)

      Here
      At \( x=\frac{1}{2}^{-}\), the left hand limit is
      LHL=\( \displaystyle \lim_{x \to \frac{1}{2}^{-}} f(x)= \lim_{x \to \frac{1}{2}^{-} } \frac{1}{2}+x=\) 1
      At \( x=\frac{1}{2}^{+}\), the right hand limit is
      RHL=\( \displaystyle \lim_{x \to \frac{1}{2}^{+}} f(x)= \lim_{x \to \frac{1}{2}^{+}} \frac{3}{2}-x=\) 1
      At \(x=\frac{1}{2}\), the value of the function is
      F=\(f(\frac{1}{2}) =\) \(\frac{1}{2}\)
      The function f(x) is discontinuous at \(x=\frac{1}{2}\).

      Since, LHL=RHL ≠f(x), the f(x) has removable discontinuity at \(x=\frac{1}{2}\)
   2. A function f(x) is defined in (0,3) as follows
      \(f(x)=\begin{cases} x^2 & \text {for } 0 < x <1 \\ x & \text{for } 1 \le x < 2 \\ \frac{1}{4}x^3 & \text{for } 2 \le x < 3 \end{cases}\)
      Show that f(x) is continuous at x=1 and x=2

      Solution 👉 Click Here

      Given \(f(x)=\begin{cases} x^2 & \text {for } 0 < x <1 \\ x & \text{for } 1 \le x < 2 \\ \frac{1}{4}x^3 & \text{for } 2 \le x < 3 \end{cases}\)

      test at x=1

      Here
      At \( x=1^{-}\), the left hand limit is
      LHL=\( \displaystyle \lim_{x \to 1^{-}} f(x)= \lim_{x \to 1^{-} } x^2=\) 1
      At \( x=1^{+}\), the right hand limit is
      RHL=\( \displaystyle \lim_{x \to 1^{+}} f(x)= \lim_{x \to 1^{+}} x=\) 1
      At \(x=1\), the value of the function is
      F=\(f(1) =x=\) 1
      Since, LHL=RHL=f(1), the function is continuous at x=1

      test at x=2

      Here
      At \( x=2^{-}\), the left hand limit is
      LHL=\( \displaystyle \lim_{x \to 2^{-}} f(x)= \lim_{x \to 2^{-} } x=\) 2
      At \( x=2^{+}\), the right hand limit is
      RHL=\( \displaystyle \lim_{x \to 2^{+}} f(x)= \lim_{x \to 2^{+}} \frac{1}{4}x^3=\) 2
      At \(x=2\), the value of the function is
      F=\(f(2) =\frac{1}{4}x^3=\) 2
      Since, LHL=RHL=f(2), the function is continuous at x=2

   3. A function f(x) is defined as follows
      \(f(x)=\begin{cases} 3+2x & \text {for } -\frac{3}{2} \le x <0 \\ 3-2x & \text{for } 0 \le x < \frac{3}{2} \\ -3-2x & \text{for } x \ge \frac{3}{2} \end{cases}\)
      Show that f(x) is continuous at x=0 and discontinuous at \(x=\frac{3}{2}\)

      Solution 👉 Click Here

      Given \(f(x)=\begin{cases} 3+2x & \text {for } -\frac{3}{2} \le x <0 \\ 3-2x & \text{for } 0 \le x < \frac{3}{2} \\ -3-2x & \text{for } x \ge \frac{3}{2} \end{cases}\)

      test at x=0

      Here
      At \( x=0^{-}\), the left hand limit is
      LHL=\( \displaystyle \lim_{x \to 0^{-}} f(x)= \lim_{x \to 0^{-} } 3+2x=\) 3
      At \( x=0^{+}\), the right hand limit is
      RHL=\( \displaystyle \lim_{x \to 0^{+}} f(x)= \lim_{x \to 0^{+}} 3-2x=\) 3
      At \(x=0\), the value of the function is
      F=\(f(0) =3-2x=\) 3
      Since, LHL=RHL=f(0), the function is continuous at x=0

      test at x=3/2

      Here
      At \( x=\frac{3}{2}^{-}\), the left hand limit is
      LHL=\( \displaystyle \lim_{x \to \frac{3}{2}^{-}} f(x)= \lim_{x \to \frac{3}{2}^{-} } 3-2x=\) 0
      At \( x=\frac{3}{2}^{+}\), the right hand limit is
      RHL=\( \displaystyle \lim_{x \to \frac{3}{2}^{+}} f(x)= \lim_{x \to \frac{3}{2}^{+}} -3-2x=\) -6
      At \(x=\frac{3}{2}\), the value of the function is
      F=\(f(\frac{3}{2}) =-3-2x=\) -6
      Since, LHL≠RHL=f(3/2), the function is NOT continuous at x=3/2

   4. A function f(x) is defined as follows
      \(f(x)=\begin{cases} 1 & \text {for } x >0 \\ 0 & \text{for } x=0 \\ -1 & \text{for } x < 01\end{cases}\)
      Show that f(x) isdiscontinuous at x=0.

      Solution 👉 Click Here

      Given \(f(x)=\begin{cases} 1 & \text {for } x >0 \\ 0 & \text{for } x=0 \\ -1 & \text{for } x < 0\end{cases}\)

      Here
      At \( x=0^{-}\), the left hand limit is
      LHL=\( \displaystyle \lim_{x \to 0^{-}} f(x)= \lim_{x \to 0^{-} } -1=\) -1
      At \( x=0^{+}\), the right hand limit is
      RHL=\( \displaystyle \lim_{x \to 0^{+}} f(x)= \lim_{x \to 0^{+}} 1=\) 1
      At \(x=0\), the value of the function is
      F=\(f(0) =0=\) 0
      Since, LHL≠RHL≠f(0), the function is NOT continuous at x=0

9. In the following, determine the value of the constant so that the given function is continuous at the point mentioned.
   
   1. \(f(x)=\begin{cases} kx^2 & \text {for } x \le 2 \\ 3 & \text{for } x > 2\end{cases}\) at x=2

      Solution 👉 Click Here

      Given that
      \(f(x)=\begin{cases} kx^2 & \text {for } x \le 2 \\ 3 & \text{for } x > 2\end{cases}\)
      
      At x=2, we compute the following
      The functional value is
      \(f(x)=kx^2=\) 4k
      
      The left hand limit is
      \( LHL=\displaystyle \lim_{x \to 2^-} f(x)= \lim_{x \to 2^-} kx^2=\) 4k
      
      The right hand limit is
      \(RHL= \displaystyle \lim_{x \to 2^+} f(x)= \lim_{x \to 2^+}3=\) 3
      
      Since,
      Function is assumed continuous at x=2, thus we must have
      LHL=RHL=f(x)
      
      or 4k=3
      or \(k=\frac{3}{4}\)
      

   2. \(f(x)=\begin{cases}ax+5 & \text {for } x \le 2 \\ x-1 & \text{for } x > 2\end{cases}\) at x=2

      Solution 👉 Click Here

      Given that
      \(f(x)=\begin{cases}ax+5 & \text {for } x \le 2 \\ x-1 & \text{for } x > 2\end{cases}\)
      
      At x=2, we compute the following
      The functional value is
      \(f(x)=ax+5=\) 2a+5
      
      The left hand limit is
      \( LHL=\displaystyle \lim_{x \to 2^-} f(x)= \lim_{x \to 2^-} ax+5=\) 2a+5
      
      The right hand limit is
      \(RHL= \displaystyle \lim_{x \to 2^+} f(x)= \lim_{x \to 2^+}x-1=\) 1
      
      Since,
      Function is assumed continuous at x=2, thus we must have
      LHL=RHL=f(x)
      
      or 2a+5=1
      or \(a=2\)
      

   3. \(f(x)=\begin{cases} 2px+3 & \text {for } x <1 \\ 1-px^2& \text{for } x \ge 1\end{cases}\) at x=1

      Solution 👉 Click Here

      Given that
      \(f(x)=\begin{cases} 2px+3 & \text {for } x <1 \\ 1-px^2& \text{for } x \ge 1\end{cases}\)
      
      At x=1, we compute the following
      The functional value is
      \(f(x)=1-px^2=\) 1-p
      
      The left hand limit is
      \( LHL=\displaystyle \lim_{x \to 1^-} f(x)= \lim_{x \to 1^-} 2px+3=\) 2p+3
      
      The right hand limit is
      \(RHL= \displaystyle \lim_{x \to 1^+} f(x)= \lim_{x \to 1^+}1-px^2=\) 1-p
      
      Since,
      Function is assumed continuous at x=1, thus we must have
      LHL=RHL=f(x)
      
      or 1-p=2p+3
      or \(p=-\frac{2}{3}\)
      

   4. \(f(x)=\begin{cases} \frac{x^2-9}{x-3} & \text {for } x \ne 3 \\ k & \text{for } x =3 \end{cases}\) at x=3

      Solution 👉 Click Here

      Given that
      \(f(x)=\begin{cases} \frac{x^2-9}{x-3} & \text {for } x \ne 3 \\ k & \text{for } x =3 \end{cases}\)
      
      At x=3, we compute the following
      The functional value is
      \(f(x)=k=\) k
      
      The left hand limit is
      \( LHL=\displaystyle \lim_{x \to 3^-} f(x)= \lim_{x \to 3^-} \frac{x^2-9}{x-3}=\lim_{x \to 3^-} (x+3)=\) 6
      
      The right hand limit is
      \(RHL= \displaystyle \lim_{x \to 3^+} f(x)= \lim_{x \to 3^+}\frac{x^2-9}{x-3}=\lim_{x \to 3^-} (x+3)=\) 6
      
      Since,
      Function is assumed continuous at x=3, thus we must have
      LHL=RHL=f(x)
      
      or k=6
      

10. Let \(f(x)=\begin{cases} 2x & \text {for } x <2 \\ 2 & \text{for } x=2 \\ x^2 & \text{for } x >2 \end{cases}\). Show that f(x) has removable discontinuity at x=2.

    Solution 👉 Click Here

    Given that
    \(f(x)=\begin{cases} 2x & \text {for } x <2 \\ 2 & \text{for } x=2 \\ x^2 & \text{for } x >2 \end{cases}\)
    
    At x=2, we compute the following
    The functional value is
    \(f(x)=2=\) 2
    
    The left hand limit is
    \( LHL=\displaystyle \lim_{x \to 2^-} f(x)= \lim_{x \to 2^-} 2x=\) 4
    
    The right hand limit is
    \(RHL= \displaystyle \lim_{x \to 2^+} f(x)= \lim_{x \to 2^+}x^2=\) 4
    
    Since,
    LHL=RHL≠f(x)
    
    The function f(x) has removable discontinuity at x=2.
    

11. What condition is necessary for a function f(x) to be continuous at the point x=a? In what condition will f(x) be discontinuous at x=a?

    Solution 👉 Click Here

    Let 𝑓(𝑥) be a function defined at all values in an open interval containing a
    Then we say that f(x) is continuous at a if
    for a given number \(\epsilon > 0\), there exists a number \(\delta > 0\)such that
    |f(x)-f(a)|<\(\epsilon\) whenever |x-a|<\(\delta\)

    OR
    

    the function f(x) is continuous at x=a if the following three conditions holds

    1. f(a) is defined
    2. \( \displaystyle \lim_{x\to a^{-}}f(x)\) exist = LHL
    3. \( \displaystyle \lim_{x\to a^{+}}f(x) \) exist = RHL
    4. LHL = RHL = f(a)

    If one of the following condition occured, then the function f(x) will be discontinuous at x=a

    1. LHL = RHL, BUT f(a) is NOT defined [Removable Discontinuity]
    2. f(a)=LHL = RHL≠f(a)[Removable Discontinuity]
    3. f(a)=LHL≠ RHL[Jump Discontinuity]
    4. LHL≠ RHL=f(a)[Jump Discontinuity]
    5. LHL= RHL=∞ or LHL= RHL=-∞[Infinite Discontinuity]
    6. LHL=∞, RHL=-∞ or LHL=-∞, RHL=∞[Infinite Discontinuity]
    7. LHL ,RHL ,f(a) are NOT equal [Discontinuity]
12. A function f(x) is defined as \(f(x)=\begin{cases} 1 & \text {for } x \ne 0 \\ 2 & \text{for } x =0 \end{cases}\). Find \(\displaystyle \lim_{x \to 0} f(x)\) if exists. Is the function continuous at x=0?

    Solution 👉 Click Here

    Given that
    \(f(x)=\begin{cases} 1 & \text {for } x \ne 0 \\ 2 & \text{for } x =0 \end{cases}\)
    
    At x=0, we compute the following
    The left hand limit is
    \( LHL=\displaystyle \lim_{x \to 0^-} f(x)= \lim_{x \to 0^-} 1=\) 1
    
    The right hand limit is
    \(RHL= \displaystyle \lim_{x \to 0^+} f(x)= \lim_{x \to 0^+}1=\) 1
    
    Since,
    LHL=RHL
    
    the limit \(\displaystyle \lim_{x \to 0} f(x)\) exists, and it is 1.
    Again
    The functional value at x=0 is
    \(f(x)=2=\) 2
    
    Since,
    LHL=RHL≠f(a)
    
    The function f(x) is NOT continuity at x=0.
    

13. Find the point of discontinuous of the following functions
    
    1. \(f(x)=\frac{x+1}{x-1}\)

       Solution 👉 Click Here

       The point of discontinuous of the function \(f(x)=\frac{x+1}{x-1}\) is \(x-1=0\)
       x-1=0
       
       or x=1
       
       

    2. \(f(x)=\frac{3x-1}{x^3-5x^2+6x}\)

       Solution 👉 Click Here

       The point of discontinuous of the function \(f(x)=\frac{3x-1}{x^3-5x^2+6x}\) is \(x^3-5x^2+6x=0\)
       \(x^3-5x^2+6x=0\)
       
       or \(x=0\) or \(x^2-5x^+6=0\)
       
       

       or \(x=0\) or \((x-2)(x-3)=0\)
       
       

       or \(x=0,2,3\)