# Examples On Continuity Set-4

**Example – 18**

Let \(f\left( x \right) = \left( {\begin{align}&{\frac{{{a^{2\left[ x \right] + \left\{ x \right\}}} - 1}}{{2\left[ x \right] + \left\{ x \right\}}}} {\qquad x \ne 0}\\&\qquad{\ell n\,a} {\qquad\qquad x = 0}\end{align}} \right)\). Evaluate the continuity of *f*(*x*) at *x* = 0.

**Solution: ** The LHL and RHL might differ due to the discontinuous nature of [ * x * ] and {*x*}. Lets determine their value:

\(\begin{align}{\rm{LHL}}\left( {{\rm{at }}x = 0} \right) &= \mathop {\lim }\limits_{h \to 0} f\left( { - h} \right)\\&= \mathop {\lim }\limits_{h \to 0} \frac{{{a^{2\left[ { - h} \right] + \left\{ { - h} \right\}}} - 1}}{{2\left[ { - h} \right] + \left\{ { - h} \right\}}}\end{align}\)

For *h* > 0, it is obvious that \(\left[ { - h} \right] = - 1\,{\rm{and }}\left\{ { - h} \right\} = 1 - h.\)

\(\begin{align}\text{Therefore,} \;{\rm{LHL}} &= \mathop {\lim }\limits_{h \to 0} \frac{{{a^{ - 2 + 1 - h}} - 1}}{{ - 2 + 1 - h}}\\&= \mathop {\lim }\limits_{h \to 0} \frac{{{a^{ - 1 - h}} - 1}}{{ - 1 - h}}\\&= \frac{{{a^{ - 1}} - 1}}{{ - 1}}\\& = {\rm{ }}1 - {a^{ - 1}}\end{align}\)

Similarly,

\(\begin{align}{\rm{RHL}} & = \mathop {\lim }\limits_{h \to 0} f\left( h \right)\\

&= \mathop {\lim }\limits_{h \to 0} \frac{{{a^{2\left[ h \right] + \left\{ h \right\}}} - 1}}{{2\left[ h \right] + \left\{ h \right\}}}\\\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{{a^h} - 1}}{h}\\\\& = {\rm{ }}ln\,\,a\end{align}\)

Since, \({\rm{LHL}} \ne {\rm{RHL}};f\left( x \right)\) is discontinuous at *x* = 0.

**Note: **

** (i) ** Some authors talk about removable and irremovable discontinuities. Let us discuss what this means:

Consider \(\begin{align}f\left( x \right) = \frac{{{x^2} - 1}}{{x - 1}},x \ne 1\end{align}\)

The LHL and RHL at *x* = 1 exist and both are equal to 2.

There is a hole in the graph at *x* =1 and therefore the function is discontinuous at *x* =1 We can, if we want to, fill this hole by redefining the function in the following manner:

\[f\left( x \right) = \left\{ {\begin{align}&{\frac{{{x^2} - 1}}{{x - 1}}} {\qquad \quad x \ne 1}\\&2 {\qquad\qquad \qquad x = 1}\end{align}} \right\}\]

The additional definition at *x* =1 fills the hole and removes the discontinuity. Hence, such a discontinuity would be called a ** re****movable discontinuity. **

By now, you should have realised that a discontinuity is removable only if the LHL and RHL are equal, since only then we can redefine *f *(*a*) to make all the three quantities equal.

If the LHL and RHL are themselves non-equal, no redefinition of *f *(*a*) could possibly make the function continuous and hence, such a discontinuity would be called irremovable. For example, *f* (*x*) = [*x*] and *f *(*x*) = {*x*} suffer from irremovable discontinuities at all integers.

**(ii) ** If * f * and *g* are two continuous functions at a point *x* = *a* (which is common to their domains), then \(f \pm g\) and *fg* will also be continuous at *x* = *a*. Furthermore, if \(g\left( a \right) \ne 0,\) then \(\begin{align}\,\frac{f}{g}\end{align}\) will also be continuous at *x* =* a*.

**(iii) ** If * g * is continuous at *x* =* a* and *f* is continuous at *x* = *g*(*a*), then *f*(*g*(*x*)) will be continuous at** *** x * =*a*.

**(iv) ** Any polynomial function is continuous for all values of * x * .

**(v) ** The functions sin * x * , cos *x* and \({{e}^{x}}\) (or\({{a}^{x}}\) ) are continuous for all values of *x*. *ln* *x* (or \({{\log }_{a}}x\)) is continuous for all *x* > 0.

**TRY YOURSELF – III**

**Q. 1 ** Find the values of * a * and *f *(0) if* f *(*x*) is continuous at *x* = 0, where

\(\begin{align}&{f\left( x \right) = \frac{{\sin 2x + a\sin x}}{{{x^3}}}} {\quad ,x \ne 0}\end{align}\)

**Q. 2 ** Find the value of * a * if *f *(*x*) is continuous at *x* = 0, where

\(f\left( x \right) = \left\{ {\begin{align}&{\frac{{1 - \cos 4x}}{{{x^2}}}} {\qquad \quad x < 0}\\&a {\qquad \qquad \qquad\quad x = 0}\\&{\frac{{\sqrt x }}{{\sqrt {16 + \sqrt x } - 4}}} {\,\,\,\,\,\,x > 0}\end{align}} \right\}\)

**Q. 3 ** If the function * f * (*x*) defined by

\(f\left( x \right) = \left\{ {\begin{align}&{\frac{{\log \left( {1 + ax} \right) - \log \left( {1 - bx} \right)}}{x}} {\qquad x \ne 0}\\&\quad\qquad \qquad k \qquad\qquad\qquad\qquad{x = 0}\end{align}} \right\}\)

is continuous at *x* = 0, find *k*.

**Q. 4 ** Let * f * (*xy*) = *f*(*x*)* f*(*y*) for every \(x,y \in \mathbb{R}.\) If *f *(*x*) is continuous at any one point *x *=* a*, then prove that *f *(*x*) is continuous for all \(x \in \mathbb{R}\sim\{ 0\} \) .

**Q. 5 ** Find the values of * a * and *b* so that *f *(*x*) is continuous at *x* = 0, where

\(f\left( x \right) = \left\{ {\begin{align}&\qquad\qquad3 {\quad\quad \quad \quad \quad x = 0}\\&{{{\left( {1 + \frac{{ax + b{x^3}}}{{{x^2}}}} \right)}^{1/x}}} {\,\,\,\,\,\,x \ne 0}\end{align}} \right\}\)

**Q. 6 ** Discuss the continuity of * f * (*x*) = [*x*] + [–*x*] at integer points.

**Q. 7 ** Discuss the continuity of * f * (*x*) in [0, 2] where

\(f\left( x \right) = \left\{ {\begin{align}&{\cos \pi x} {\qquad \qquad \quad x \le 1}\\&{\left| {2x - 3} \right|\left[ x \right]} {\qquad \quad x > 1}\end{align}} \right\}\)

**Q. 8 ** If \(f\left( x \right) = \left\{ {\begin{align}&{ - 1} {,\qquad \quad x < 0}\\&0 {,\qquad \qquad x = 0}\\&1 {,\qquad \qquad x > 0}\end{align}} \right\}{\rm{and }}\;g\left( x \right) = x\left( {1 - {x^2}} \right),\) then discuss the continuity of *f *(*g*(*x*)), and * g *(*f *(*x*)).