 # Continuity and Differentiability

## Class 12 NCERT

### NCERT

1   Prove that the function $f ( x ) = 5 x - 3$ is continuous at $x = 0, x = - 3$ and at $x = 5.$

The given function is $F(x)=5x-3$$\\ At x=0,f(0)=5*0-3=3\\ \lim\limits_{x \to 0}f(x)=\lim\limits_{x \to 0}f(5x-3)\\ =5*0-3=-3 \\ \therefore \lim\limits_{x \to 0}f(x)=f(0)$$\\$ Therefore ,$f$ is continuous at $x=0$$\\ At x=-3,f(-3) =5x(-3)-3=-18\\ \lim\limits_{x \to 3}f(x)=\lim\limits_{x \to 3}f(5x-3)=5*(-3)-3=-18\\ \therefore \lim\limits_{x \to 3}f(x)=f(-3)$$\\$ Therefore , $f$ is continuous at $x=-3$ $\\$ At $x=5,f(x) = f(5)=5*5-3=25-3=22\\ \lim\limits_{x \to 5}f(x)=\lim\limits_{x \to 5}f(5x-3)=5*5-3=-22\\ \therefore \lim\limits_{x \to 5}f(x)=f(5)$$\\ Therefore , f is continuous at x=5 2 Examine the continuity of the function f(x)=2x^2 -1 at x=3 ##### Solution : The given function is f(x) =2x^2-1 At x=3,f(x)=f(3)=2*3^2-1=17\\ \lim\limits _{x \to 3}f{x}=\lim\limits_{x\to 3}(2x^2-1) =2*362-1=17\\ \therefore \lim\limits_{x\to 3}f(x)=f=(3)$$\\$ Thus,$f$ is continuous , at $x=3$

3   Examine the following functions for continuity.$\\$ $a) f(x) =x-5 \\ b) f(x)=\dfrac{1}{x-5},x\neq 5\\ c) f(x) =\dfrac{x^2-25}{x+5},x\neq 5\\ d) f(x) = |x-5|$

##### Solution :

The given function is $f(x) =|x-5|=\begin{cases} 5-x & \quad \text{if } x<5\\ x-5 & \quad \text{if } x\geq 5 \end{cases}$$\\ This function f is defined at all points of the real line.\\ Let c be a point on a real line. Then,c<5 O r c=5 Or c>5$$\\$ case $I:c<5$$\\ Then, f(c)=5-c\\ \lim\limits_{x \to c}f(x)=\lim\limits_{x\to c}(5-x)=5-c\\ \therefore \lim\limits_{x \to c} f(x)=f(c)$$\\$ Therefore,$f$ is continuous at all real numbers less than $5$.$\\$ case $II: c=5$$\\ Then , f(c)=f(5)=(5-5)=0\\ \lim\limits_{x \to 5^-}f(x)=\lim\limits_{x \to 5} (5-x)=(5-5)=0\\ \lim\limits_{x \to 5^+}f(x)=\lim\limits_{x \to 5}(x-5)=0\\ \therefore \lim\limits_{x \to c^-}f(x)=\lim\limits_{x\to c^+}f(x)=f(c)$$\\$ Therefore ,$f$ is contitinuous at $x=5$$\\ a) The given function is f (x) = x - 5$$\\$ It is evident that $f$ is defined at every real number k and its value at $k$ is $k -5 .$$\\ It is also observed that \lim\limits _{x \to k}f(x)=\lim\limits_{x \to k}(x-5)=k=k-5=f(k)\\ \therefore \lim\limits _{x\to k}f(x)=f(k)$$\\$ Hence ,$f$ is continuous at every real number and therefore, it is a continuous function.

$c)$ The given function is $f(x) =\dfrac{x^2-25}{x+5},x\neq -5$$\\ For any real number c \neq - 5 , we obtain \\ \lim\limits_{x \to c}(x) =\lim\limits_{x \to c}(\dfrac{x62-25}{x+5})=\lim\limits_{x \to c}\dfrac{(x+5)(x-5)}{x+5} \\ =\lim\limits_{x \to c}(x-5) =(c-5)$$\\$ Also ,$f(c) =\dfrac{(c+5)(c-5)}{c+5} =c(c-5)(as c \neq 5)\\ \therefore \lim\limits_{x\to c}f(x)=f(c)$$\\ Hence f is continuous at every point in the domain of f and therefore. It is continuous function. case III: c>5$$\\$ Then,$f(c)=f(5) =c-5\\ \lim\limits_{x\to c}f(x)=\lim\limits_{x \to c}f(x-5)=c-5\\ \therefore \lim\limits_{x \to c}f(x)=f(c)$$\\ Therefore, f is continuous at real numbers greater than 5.\\ Hence, f is continuous at every real number and therefore, it is a continuous function. b) The given function is f(x)=\dfrac{1}{x-5},x \neq 5$$\\$ for any real number $k \neq 5$ , we obtain$\\$ $\lim\limits _{x \to k}f(x)=\lim\limits_{x\to k}\dfrac{1}{x-5}=\dfrac{1}{k-5}$$\\ Also,f(k) =\dfrac{1}{k-5} \ \ \ \ \ \ \ \ \ \ \ \ \ (As k\neq 5)\\ \therefore \lim\limits _{x\to k}f(x)=f(k)$$\\$ Hence, $f$ is continuous at every point in the domain of $f$ and therefore, it is a continuous function.

4   Prove that the function $f(x)=x^n$ is continuous at $x=n$ is a positive integer.

##### Solution :

The given function is $f (x ) =x^ n$ It is evident that $f$ is defined at all positive integers, $n$ , and its value at $n$ is $n^ n$ .$\\$ Then,$\lim\limits_{x\to n}f(n)=\lim\limits_{x \to n}f(x^n)=n^n\\ \therefore \lim\limits_{x \to n}f(x)=f(n)$$\\ Therefore, f is continuous at n , where n is a positive integer. 5 Is the function f defined by f(x)=\begin{cases} x & \quad \text{if } x\leq\\ 5 & \quad \text{if } x>1 \end{cases}$$\\$ Continuous at $x=0?$ At $x=2?$

##### Solution :

The given function $f$ is $f(x)=\begin{cases} x & \quad \text{if } x\leq\\ 5 & \quad \text{if } x>1 \end{cases}$$\\ At x = 0,$$\\$ It is evident that $f$ is defined at $0$ and its value at $0$ is $0$ . Then,$\lim\limits_{x \to 0}f(x)=\lim\limits_{x \to 0}x=0 \\ \therefore \lim\limits_{x \to 0}f(x) =f(0)$$\\ Therefore, f is continuous at x = 0$$\\$ At $x = 1 ,$$\\ f is defined at 1 and its value at is 1 . The left hand limit of f at x = 1 is,\\ \lim\limits_{x \to 1^-}f(x)=\lim\limits_{x \to 1^-} x=1$$\\$ The right hand limit of $f$ at $x=1$ is ,$\\$ $\lim\limits_{x \to 1 ^+}f(x) =\lim\limits_{x \to 1 ^+}f(5)\\ \therefore \lim\limits_{x \to 1^-}f(x)\neq \lim\limits_{x \to 1^+}f(x)$$\\ Therefore, f is not continuous at x = 1$$\\$ At $x = 2,$$\\ f is defined at 2 and its value at 2 is 5.\\ Then,\lim\limits_{x\to 2}f(x)=\lim\limits_{x \to 2}f(5)=5\\ \therefore \lim\limits_{x \to 2}f(x)=f(2)$$\\$ Therefore,$f$ is continuous at $x = 2$

6   Find all points of discontinuous of $f$ , where $f$ is defined by$\\$ $f(x)= \begin{cases} 2x+3 & \quad \text{if } x \leq 2\\ 2x-3 & \quad \text{if }x >2 \end{cases}$

##### Solution :

The give function $f$ is $f(x)= \begin{cases} 2x+3 & \quad \text{if } x \leq 2\\ 2x-3 & \quad \text{if }x >2 \end{cases}$$\\ It is evident that the given function f is defined at all the points of the real line. Let c be a point on the real line. Then, three cases arise.\\ I. c < 2\\ II. c > 2\\ III. c=2$$\\$ Case(i)$c < 2$$\\ Then, f(x)=2x+3\\ \lim\limits_{x \to c} f(x) =\lim\limits_{x \to c}(2x+3)=2c+3\\ \therefore \lim\limits_{x \to c} f(x)=f(c)$$\\$ Therefore, $f$ is continuous at all points, $x$ , such that $x < 2$$\\ Case(ii)c=2$$\\$ Then,$f(c)=2c-3\\ \lim\limits_{x \to c} f(x)=\lim\limits_{x \to c}(2x-3)=2c-3\\ \therefore \lim\limits{x \to c} f(x) =f(c)$$\\ Therefore, f is continuous at all points x , such that x > 2$$\\$ Case(iii)$c=2 $$\\ Then, the left hand limit of f at x =2 is,\\ \lim\limits_{x \to 2^-} f(x)=\lim\limits_{x \to 2^-}(2x+3)=2*2+3=7$$\\$ The right hand limit of $f$ at $x = 2$ is,$\\$ $\lim\limits_{x \to 2^+} f(x)=\lim\limits{x \to 2^+}(2x+3)=2*2-3=1$$\\ It is observed that the left and right hand limit of f at x = 2 do not coincide. Therefore, f is not continuous at x = 2$$\\$ Hence, $x = 2$ is the only point of discontinuity of $f$ .

7   Find all points of discontinuity of $f$ , where $f$ is defined by$\\$ $f(x)=\begin{cases} |x|+3 & \quad \text{if } x\leq -3\\ -2x & \quad \text{if } -3 < x < 3\\ 6x+2& \quad \text{if} x \geq 3 \end{cases}$

##### Solution :

If$-3 < c < 3$,then $f(c)=-2c$ and$\\$ $\lim\limits_{x \to c} f(x)=\lim\limits_{x \to 3c}(-2x)=-2c\\ \therefore \lim\limits_{x \to c} f(x)=f(c)$$\\ Therefore, f is continuous in (- 3,3 ) .\\ Case IV :\\ If c = 3, then the left hand limit of f at x = 3 is,\\ \lim\limits_{x \to 3^-} f(x)=\lim\limits_{x \to 3^-} f(-2x)=-2*3=6$$\\$ The right hand limit of $f$ at $x =3$ is,$\\$ $\lim\limits_{x \to 3^+} f(x) =\lim\limits_{x \to 3^+} f(6x+2)=6*3+2=20$$\\ It is observed that the left and right hand limit of f at x = 3 do not coincide. Therefore, f is not continuous at x = 3$$\\$ Case V :$\\$ If $c > 3$ , then $f ( c)=6 c + 2$ and $\\$ $\lim\limits_{x \to c} f(x) =\lim\limits_{x \to c} f(6x+2)=6c+2\\ \therefore \lim\limits_{x \to c} f(x)= f(c)$$\\ Therefore, f is continuous at all points x , such that x > 3$$\\$ Hence, $x = 3$ is the only point of discontinuity of $f$ .

The given function $f$ is $f(x)=\begin{cases} |x|+3 & \quad \text{if } x\leq -3\\ -2x & \quad \text{if } -3 < x < 3\\ 6x+2& \quad \text{if} x \geq 3 \end{cases}$$\\ The given function f is defined at all the points of the real line. Let c be a point on the real line.\\ Case I.\\ If c < -3, then f(c)=-c+3$$\\$ $\lim\limits_{x \to c} f(x)=\lim\limits_{x \to c}(-x+3)=-c+3\\ \therefore \lim\limits_{x \to c} f(x)=f(c)$$\\ Therefore, f is continuous at all points x , such that x <- 3$$\\$ Case II.$\\$ If$c=-3,$ then $f(-3)=-(-3)+3=6\\ \lim\limits_{x \to 3^-} f(x)=\lim\limits_{x \to 3^-}(-x+3)=-(-3)+3=6\\ \therefore \lim\limits_{x \to 3^+} f(x) =\lim\limits_{x \to 3^+} f(-2x) =2x(-3)=6\\ \therefore \lim\limits_{x \to 3 } f(x)=f(-3)$$\\ Therefore, f is continuous at x=- 3$$\\$ Case III.$\\$If$-3 < c < 3$,then $f(c)=-2c$ and$\\$ $\lim\limits_{x \to c} f(x)=\lim\limits_{x \to 3c}(-2x)=-2c\\ \therefore \lim\limits_{x \to c} f(x)=f(c)$$\\ Therefore, f is continuous in (- 3,3 ) .\\ Case IV :\\ If c = 3, then the left hand limit of f at x = 3 is,\\ \lim\limits_{x \to 3^-} f(x)=\lim\limits_{x \to 3^-} f(-2x)=-2*3=6$$\\$ The right hand limit of $f$ at $x =3$ is,$\\$ $\lim\limits_{x \to 3^+} f(x) =\lim\limits_{x \to 3^+} f(6x+2)=6*3+2=20$$\\ It is observed that the left and right hand limit of f at x = 3 do not coincide. Therefore, f is not continuous at x = 3$$\\$ Case V :$\\$ If $c > 3$ , then $f ( c)=6 c + 2$ and $\\$ $\lim\limits_{x \to c} f(x) =\lim\limits_{x \to c} f(6x+2)=6c+2\\ \therefore \lim\limits_{x \to c} f(x)= f(c)$$\\ Therefore, f is continuous at all points x , such that x > 3$$\\$ Hence, $x = 3$ is the only point of discontinuity of $f$ .

8   Find all points of discontinuity of $f$ , where $f$ is defined by $f ( x )= \begin{cases} |x| & \quad \text{if } x\neq 0\\ x \\ 0 & \quad \text{if } x=0 \end{cases}$

##### Solution :

The given function $f$, where $f$ is defind by $f(x)=\begin{cases} |x| & \quad \text{if } x\neq 0\\ x \\ 0 & \quad \text{if } x=0 \end{cases}$$\\ It is known that,x < 0 \Rightarrow |x| =-x and x > 0 \Rightarrow |x| =x$$\\$ Therefore, the given function can be rewritten as $\\$ $f(x)= \begin{cases} \dfrac{|x|}{x}=\dfrac{-x}{x}=-1 & \quad \text{if } x < 0\\ 0, & \quad \text{if} x=0\\ \dfrac{|x|}{x}=\dfrac{x}{x}=1 & \quad \text{if } x > 0 \end{cases}$$\\ The given function f is defined at all the points of the real line.\\ Let c be a point on the real line. Case I :\\ if c < 0, then f(c)=-1$$\\$ $\lim\limits_{x \to c}f(x)=\lim\limits_{x \to c}(-1)=-1\\ \therefore \lim\limits_{x \to c} f(x) =f(c)$$\\ Therefore,f is continuous at all points x < 0$$\\$ Case II.$\\$ If $c=0,$ then the left hand limit of $f$ at $x = 0$ is,$\\$ $\lim\limits_{x \to 0^-} f(x)= \lim\limits_{x \to 0^-}(-1)=-1$$\\ The right hand limit of f at x = 0 is,\\ Case III. \\ If c > 0,f(c)=1$$\\$ $\lim\limits_{x \to c}f(x)=\lim\limits_{x \to c} (1) =1\\ \therefore \lim\limits_{x \to c}(x)=f(c)$$\\ Therefore, f is continuous at all points x , such that x > 0$$\\$ Hence, $x = 0$ is the only point of discontinuity of $f$ .

9   Find all points of discontinuity of $f$, where $f$ is defined by $f (x)=\begin{cases} \dfrac{x}{|x|} & \quad \text{if }x < 0\\ -1 & \quad \text{if } x \geq 0 \end{cases}$