Q: Is the function $f$ defined by $f(x)=\left\{\begin{array}{l}x, \text { if } x \leq 1 \\ 5, \text { if } x>1\end{array}\right.$ continuous at $x=0$ ? At $x=1$ ? At $x=2$ ?

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

Q: Find all points of discontinuity of $f$, where $f$ is defined by $f(x)=\left\{\begin{array}{l}2 x+3, \text { if } x \leq 2 \\ 2 x-3, \text { if } x>2 .\end{array}\right.$

The given function is $f(x)=\left\{\begin{array}{l}2 x+3, \text { if } x \leq 2 \\ 2 x-3, \text { if } x>2\end{array}\right.$ 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. $c2$ $c=2$ Case I: $c2$ Then, $f(c)=2 c-3$ $\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(2 x-3)=2 c-3$ $\therefore \lim _{x \rightarrow c} f(x)=f(c)$ 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, $\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{-}}(2 x+3)=2(2)+3=7$ The right hand limit of $f$ at $x=2$ is, $$ \lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{+}}(2 x-3)=2(2)-3=1 $$ 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 $f$.

Q: Find all points of discontinuity of $f$, where $f$ is defined by $$ f(x)=\left\{\begin{array}{l} |x|+3, \text { if } x \leq-3 \\ -2 x, \text { if }-3<x<3 \\ 6 x+2, \text { if } x \geq 3 \end{array}\right. $$

The given function is $f(x)=\left\{\begin{array}{l}|x|+3, \text { if } x \leq-3 \\ -2 x, \text { if }-33$, then $f(c)=6 c+2$ $\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(6 x+2)=6 c+2$ $\therefore \lim _{x \rightarrow c} f(x)=f(c)$ Therefore, $f$ is continuous at all points $x$, such that $x>3$. Hence, $x=3$ is the only point of discontinuity of $f$.

Question 1

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

Accepted Answer

The given function is $f(x)=5 x-3$

At $x=0, f(0)=5(0)-3=-3$

$\lim _{x ightarrow 0} f(x)=\lim _{x ightarrow 0}(5 x-3)=5(0)-3=-3$

$	herefore \lim _{x ightarrow 0} f(x)=f(0)$

Therefore, $f$ is continous at $x=0$.

At $x=-3, f(-3)=5(-3)-3=-18$

$\lim _{x ightarrow-3} f(x)=\lim _{x ightarrow-3}(5 x-3)=5(-3)-3=-18$

$	herefore \lim _{x ightarrow-3} f(x)=f(-3)$

Therefore, $f$ is continous at $x=-3$.

At $x=5, f(5)=5(5)-3=22$

$\lim _{x ightarrow 5} f(x)=\lim _{x ightarrow 5}(5 x-3)=5(5)-3=22$

$	herefore \lim _{x ightarrow 5} f(x)=f(5)$

Therefore, $f$ is continous at $x=5$.

Question 2

Examine the continuity of the function $f(x)=2 x^{2}-1$ at $x=3$.

Accepted Answer

The given function is $f(x)=2 x^{2}-1$

At $x=3, f(3)=2(3)^{2}-1=17$

$\lim _{x ightarrow 3} f(x)=\lim _{x ightarrow 3}\left(2 x^{2}-1ight)=2\left(3^{2}ight)-1=17$

$	herefore \lim _{x ightarrow 3} f(x)=f(3)$

Therefore, $f$ is continous at $x=3$.

Question 3

Examine the following functions for continuity.

(i)

$$

f(x)=x-5

$$

(ii) $f(x)=\frac{1}{x-5}, x 
eq 5$

(iii) $f(x)=\frac{x^{2}-25}{x+5}, x 
eq-5$

(iv) $f(x)=|x-5|, x 
eq 5$

Accepted Answer

(i) 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

$$

\begin{aligned}

& \lim _{x ightarrow k} f(x)=\lim _{x ightarrow k}(x-5)=k-5=f(k) \

& 	herefore \lim _{x ightarrow k} f(x)=f(k)

\end{aligned}

$$

Hence, $f$ is continuous at every real number and therefore, it is a continuous function.

(ii) The given function is $f(x)=\frac{1}{x-5}, x 
eq 5$

For any real number $k 
eq 5$, we obtain

$$

\lim _{x ightarrow k} f(x)=\lim _{x ightarrow k} \frac{1}{x-5}=\frac{1}{k-5}

$$

Also,

$$

\begin{aligned}

& f(k)=\frac{1}{k-5} \quad(	ext { As } k 
eq 5) \

& 	herefore \lim _{x ightarrow k} f(x)=f(k)

\end{aligned}

$$

Hence, $f$ is continuous at every point in the domain of $f$ and therefore, it is a continuous function.

(iii) The given function is $f(x)=\frac{x^{2}-25}{x+5}, x 
eq-5$

For any real number $c 
eq-5$, we obtain

$$

\lim _{x ightarrow c} f(x)=\lim _{x ightarrow c} \frac{x^{2}-25}{x+5}=\lim _{x ightarrow c} \frac{(x+5)(x-5)}{x+5}=\lim _{x ightarrow c}(x-5)=(c-5)

$$

Also,

$$

\begin{aligned}

& f(c)=\frac{(c+5)(c-5)}{c+5}=(c-5) \

& 	herefore \lim _{x ightarrow c} f(x)=f(c)

\end{aligned}

$$

Hence, $f$ is continuous at every point in the domain of $f$ and therefore, it is a continuous function.

(iv) The given function is $f(x)=|x-5|=\left\{\begin{array}{l}5-x, 	ext { if } x5$

Case I: $c5$

Then, $f(c)=f(5)=c-5$

$\lim _{x ightarrow c} f(x)=\lim _{x ightarrow c}(x-5)=c-5$

$	herefore \lim _{x ightarrow c} f(x)=f(c)$

Therefore, $f$ is continuous at all real numbers greater than 5 .

Hence, $f$ is continuous at every real number and therefore, it is a continuous function.

Question 4

Prove that the function $f(x)=x^{n}$ is continuous at $x=n$, where $n$ is a positive integer.

Accepted Answer

The given function is $f(x)=x^{n}$

It is observed that $f$ is defined at all positive integers, $n$, and its value at $n$ is $n^{n}$.

Then,

$\lim _{x ightarrow n} f(n)=\lim _{x ightarrow n}\left(x^{n}ight)=x^{n}$

$	herefore \lim _{x ightarrow n} f(x)=f(n)$

Therefore, $f$ is continuous at $n$, where $n$ is a positive integer.

Question 5

Is the function $f$ defined by $f(x)=\left\{\begin{array}{l}x, 	ext { if } x \leq 1 \ 5, 	ext { if } x>1\end{array}ight.$ continuous at $x=0$ ? At $x=1$ ? At $x=2$ ?

Accepted Answer

The given function is $f(x)=\left\{\begin{array}{l}x, 	ext { if } x \leq 1 \ 5, 	ext { if } x>1\end{array}ight.$

At $x=0$,

It is evident that $f$ is defined at 0 and its value at 0 is 0 .

Then,

$\lim _{x ightarrow 0} f(x)=\lim _{x ightarrow 0}(x)=0$

$	herefore \lim _{x ightarrow 0} f(x)=f(0)$

Therefore, $f$ is continuous at $x=0$.

At $x=1$,

It is evident that $f$ is defined at 1 and its value at 1 is 1 .

The left hand limit of $f$ at $x=1$ is,

$\lim _{x ightarrow 1^{-}} f(x)=\lim _{x ightarrow 1^{-}}(x)=1$

The right hand limit of $f$ at $x=1$ is,

$\lim _{x ightarrow 1^{+}} f(x)=\lim _{x ightarrow 1^{+}}(5)=5$

$	herefore \lim _{x ightarrow 1^{-}} f(x) 
eq \lim _{x ightarrow 1^{+}} f(x)$

Therefore, $f$ is not continuous at $x=1$.

At $x=2$,

It is evident that $f$ is defined at 2 and its value at 2 is 5 .

$\lim _{x ightarrow 2} f(x)=\lim _{x ightarrow 2}(5)=5$

$	herefore \lim _{x ightarrow 1} f(x)=f(2)$

Therefore, $f$ is continuous at $x=2$.

Question 6

Find all points of discontinuity of $f$, where $f$ is defined by $f(x)=\left\{\begin{array}{l}2 x+3, 	ext { if } x \leq 2 \ 2 x-3, 	ext { if } x>2 .\end{array}ight.$

Accepted Answer

The given function is $f(x)=\left\{\begin{array}{l}2 x+3, 	ext { if } x \leq 2 \ 2 x-3, 	ext { if } x>2\end{array}ight.$

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.

$c2$

$c=2$

Case I: $c2$

Then,

$f(c)=2 c-3$

$\lim _{x ightarrow c} f(x)=\lim _{x ightarrow c}(2 x-3)=2 c-3$

$	herefore \lim _{x ightarrow 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,

$\lim _{x ightarrow 2^{-}} f(x)=\lim _{x ightarrow 2^{-}}(2 x+3)=2(2)+3=7$

The right hand limit of $f$ at $x=2$ is,

$$

\lim _{x ightarrow 2^{+}} f(x)=\lim _{x ightarrow 2^{+}}(2 x-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 $f$.

Question 7

Find all points of discontinuity of $f$, where $f$ is defined by

$$

f(x)=\left\{\begin{array}{l}

|x|+3, 	ext { if } x \leq-3 \

-2 x, 	ext { if }-3<x<3 \

6 x+2, 	ext { if } x \geq 3

\end{array}ight.

$$

Accepted Answer

The given function is $f(x)=\left\{\begin{array}{l}|x|+3, 	ext { if } x \leq-3 \ -2 x, 	ext { if }-33$, then $f(c)=6 c+2$

$\lim _{x ightarrow c} f(x)=\lim _{x ightarrow c}(6 x+2)=6 c+2$

$	herefore \lim _{x ightarrow 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 $f$.

Question 8

Find all points of discontinuity of $f$, where $f$ is defined by $f(x)=\left\{\begin{array}{l}\frac{|x|}{x} 	ext {, if } x 
eq 0 \ 0 	ext {, if } x=0\end{array}ight.$

Accepted Answer

The given function is $f(x)=\left\{\begin{array}{l}\frac{|x|}{x}, 	ext { if } x 
eq 0 \ 0, 	ext { if } x=0\end{array}ight.$

It is known that, $x0 \Rightarrow|x|=x$

Therefore, the given function can be rewritten as

$f(x)=\left\{\begin{array}{l}\frac{|x|}{x}=\frac{-x}{x}=-1, 	ext { if } x0\end{array}ight.$

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 $c0$, then $f(c)=1$

$\lim _{x ightarrow c} f(x)=\lim _{x ightarrow c}(1)=1$

$	herefore \lim _{x ightarrow c} f(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 $f$.

5. Continuity & Differentiability

5.1

5.2

5.3

5.4

5.5

5.6

5.7

Miscellaneous