Question

(i) Examine the function for continuity.
$f\left(x\right)=x-5$

(ii) Examine the function for continuity.
$f\left(x\right)=\frac{1}{x-5},x\ne 5$

(iii)Examine the function for continuity.
$f\left(x\right)=\frac{x^2-25}{x+5},x\ne -5$

(iv)Examine the function for continuity.
$f\left(x\right)=|x-5|$

Answer 1

(i) Given function is $f\left(x\right)=x-5$
Now we check the continuity at $x=a$
( $a$ is any real constant)

If $f\left(x\right)$ is continuous at $x=a$ then

$\underset{x\to a^{-}}{lim}f\left(x\right)=\underset{x\to a^{+}}{lim}f\left(x\right)=f\left(a\right)$

Finding L.H.L.

$\underset{x\to a^{-}}{lim}x-5$

$=\underset{h\to 0}{lim}(a-h)-5$

Putting $h=0$ then we get,
$=(a-0)-5=a-5$

Finding R.H.L.

$\underset{x\to a^{+}}{lim}x-5$

$=\underset{h\to 0}{lim}(a+h)-5$

Putting $h=0$ then we get,
$=(a+h)-5=a-5$

To find $f\left(x\right)$ at $x=a$
$f\left(x\right)=x-5\text{at}x=a$
$f\left(a\right)=a-5$

Hence, $\underset{x\to a^{-}}{lim}f\left(x\right)=\underset{x\to a^{+}}{lim}f\left(x\right)=f\left(a\right)$

Therefore, the function $f\left(x\right)=x-5$ is continuous everywhere.

(ii) Given function is $f\left(x\right)=\frac{1}{x-5},x\ne 5$

Since the given function $f\left(x\right)=\frac{1}{x-5},x\ne 5$

is not defined at $x=5$ so, we don’t check continuity at $x=5$
Now we check the continuity at $x=a$
($a$ is any real constant except $5$)

If $f\left(x\right)$ is continuous at $x=a$ then

$\underset{x\to a^{-}}{lim}f\left(x\right)=\underset{x\to a^{+}}{lim}f\left(x\right)=f\left(a\right)$

Finding L.H.L.

$\underset{x\to a^{-}}{lim}\frac{1}{x-5}$

$=\underset{h\to 0}{lim}\frac{1}{(a-h)-5}$

Putting $h=0$ then we get,
$=\frac{1}{(a-0)-5}=\frac{1}{a-5}$

Finding R.H.L.

$\underset{x\to a^{+}}{lim}\frac{1}{x-5}$

$=\underset{h\to 0}{lim}\frac{1}{(a+h)-5}$

Putting $h=0$ then we get,
$=\frac{1}{(a+0)-5}=\frac{1}{a-5}$

To find $f\left(x\right)$ at $x=a$
$f\left(x\right)=\frac{1}{a-5}\text{at}x=a$
$f\left(a\right)=\frac{1}{a-5}$

Hence, $\underset{x\to a^{-}}{lim}f\left(x\right)=\underset{x\to a^{+}}{lim}f\left(x\right)=f\left(a\right)$

Therefore, the function $f\left(x\right)=\frac{1}{x-5}$, $x\ne 5$ is continuous in its domain i.e., $R-\left\{5\right\}$

(iii) Given function is $f\left(x\right)=\frac{x^2-25}{x+5},x\ne -5$

Since the given function is not defined at $x=-5$ so, we don’t check continuity at $x=-5$
Now we check the continuity at $x=a$
($a$ is any real constant except $-5$)

If $f\left(x\right)$ is continuous at $x=a$ then

$\underset{x\to a^{-}}{lim}f\left(x\right)=\underset{x\to a^{+}}{lim}f\left(x\right)=f\left(a\right)$

Finding L.H.L.

$\underset{x\to a^{-}}{lim}\frac{x^2-25}{x+5}$

$=\underset{h\to 0}{lim}\frac{(a-h{)}^2-25}{(a-h)+5}$

Putting $h=0$ then we get,
$=\frac{(a-0{)}^2-25}{(a-0)+5}=\frac{a^2-25}{a+5}=\frac{(a+5)(a-5)}{(a+5)}$

$=(a-5)$

Finding R.H.L.

$\underset{x\to a^{-}}{lim}\frac{x^2-25}{x+5}$

$=\underset{h\to 0}{lim}\frac{(a+h{)}^2-25}{(a-h)+5}$

Putting $h=0$ then we get,
$=\frac{(a+0{)}^2-25}{(a+0)+5}=\frac{a^2+25}{a+5}=\frac{(a+5)(a-5)}{(a+5)}$

$=(a-5)$

To find $f\left(x\right)$ at $x=a$
$f\left(x\right)=\frac{x^2-25}{x+5}\text{at}x=a$

$f\left(a\right)=\frac{a^2-25}{a+5}$

$=\frac{(a+5)(a-5)}{a+5}=(a-5)$

Hence, $\underset{x\to a^{-}}{lim}f\left(x\right)=\underset{x\to a^{+}}{lim}f\left(x\right)=f\left(a\right)$

Therefore, the function $f\left(x\right)=\frac{x^2-25}{x+5}$, $x\ne -5$ is continuous in its domain i.e., $R-\{-5\}$.

(iv) Given: function is ​​​​​​​$f\left(x\right)=|x-5|$

We define the function
$f\left(x\right)=\{\begin{array}{cc}(x-5),& x-5\ge 0\\ -(x-5),& x-5<0\end{array}$

$f\left(x\right)=\{\begin{array}{cc}(x-5),& x\ge 5\\ -x+5,& x<5\end{array}$
​​​​​​​When $x=5$

If $f\left(x\right)$ is continuous at $x=5$ then

$\underset{x\to 5^{-}}{lim}f\left(x\right)=\underset{x\to 5^{+}}{lim}f\left(x\right)=f\left(5\right)$

Finding L.H.L.

$\underset{x\to 5^{-}}{lim}-x+5$

$=\underset{h\to 0}{lim}-(5-h)+5$
$=\underset{h\to 0}{lim}h$

Putting $h=0$ then we get,
$=0$

Finding R.H.L.

$\underset{x\to 5^{+}}{lim}(x-5)$

$=\underset{h\to 0}{lim}(5+h)$
$=\underset{h\to 0}{lim}h$

Putting $h=0$ then we get,
$=0$

To find $f\left(x\right)$ at $x=5$
$f\left(x\right)=(x-5)\text{at}x=5$
$f\left(5\right)=(5-5)=0$

Hence, $\underset{x\to 5^{-}}{lim}f\left(x\right)=\underset{x\to 5^{+}}{lim}f\left(x\right)=f\left(5\right)$

Therefore, the function $f\left(x\right)=|x-5|$ is continuous at $x=5$

When $x<5$
For $x<5,f\left(x\right)=-(x-5)$

Since the function $f\left(x\right)=-(x-5)$ is a polynomial so it is continuous.
$\therefore f\left(x\right)$ is continuous for $x<5$

When $x>5$
For $x>5,f\left(x\right)=(x-5)$

Since the function $f\left(x\right)=x-5$ is a polynomial so it is continuous.
$\therefore f\left(x\right)$ is continuous for $x>5$

Therefore, the function $f\left(x\right)=|x-5|$ is continuous at every real number.