# 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 :

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)$$\\ Hence, f is continuous at every point in the domain of f and therefore, it is a continuous function. The given function is f(x) =|x-5|=\begin{cases} 5-x & \quad \text{if } x<5\\ x-5 & \quad \text{if } x\geq 5 \end{cases}$$\\$ 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)$$\\ Therefore , f is contitinuous at x=5$$\\$

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 $x = 2$