NCERT Solutions · Class 12 Mathematics Continuity and Differentiability

Definition: \(\displaystyle f\) is continuous at \(\displaystyle x=c\) if \(\displaystyle f(c)\) is defined, \(\displaystyle \lim_{x\to c}f(x)\) exists, and \(\displaystyle \lim_{x\to c}f(x)=f(c)\).Here \(\displaystyle f(x)=5x-3\) is defined for every real \(\displaystyle x\).At \(\displaystyle x=0\): \(\displaystyle f(0)=5(0)-3=-3\), and by the algebra of limits \(\displaystyle \lim_{x\to 0}(5x-3)=5(0)-3=-3\). Since \(\displaystyle \lim_{x\to 0}f(x)=f(0)\), \(\displaystyle f\) is continuous at \(\displaystyle x=0\).At \(\displaystyle x=-3\): \(\displaystyle f(-3)=5(-3)-3=-18\), and \(\displaystyle \lim_{x\to -3}(5x-3)=5(-3)-3=-18\). The two agree, so \(\displaystyle f\) is continuous at \(\displaystyle x=-3\).At \(\displaystyle x=5\): \(\displaystyle f(5)=5(5)-3=22\), and \(\displaystyle \lim_{x\to 5}(5x-3)=5(5)-3=22\). The two agree, so \(\displaystyle f\) is continuous at \(\displaystyle x=5\).Hence \(\displaystyle f(x)=5x-3\) is continuous at \(\displaystyle x=0\), at \(\displaystyle x=-3\) and at \(\displaystyle x=5\).

Test the definition of continuity at \(\displaystyle x=3\): compare \(\displaystyle \lim_{x\to 3}f(x)\) with \(\displaystyle f(3)\).The function \(\displaystyle f(x)=2x^{2}-1\) is defined for all real \(\displaystyle x\), so \(\displaystyle f(3)=2(3)^{2}-1=18-1=17\).By the algebra of limits (limit of a product and of a sum), \[\lim_{x\to 3}\left(2x^{2}-1\right)=2\left(\lim_{x\to 3}x\right)^{2}-1=2(3)^{2}-1=17.\]Since \(\displaystyle \lim_{x\to 3}f(x)=17=f(3)\), the function \(\displaystyle f\) is continuous at \(\displaystyle x=3\).

Here \(\displaystyle f(x)=x^{n}\) with \(\displaystyle n\) a positive integer, defined for all real \(\displaystyle x\).Value at the point: \(\displaystyle f(n)=n^{n}\).Limit at the point: \(\displaystyle x^{n}\) is the product of \(\displaystyle n\) copies of \(\displaystyle x\), so by the product rule for limits applied \(\displaystyle n-1\) times, \[\lim_{x\to n}x^{n}=\left(\lim_{x\to n}x\right)^{n}=n^{n}.\]Since \(\displaystyle \lim_{x\to n}f(x)=n^{n}=f(n)\), the function \(\displaystyle f(x)=x^{n}\) is continuous at \(\displaystyle x=n\).(The same argument at any real \(\displaystyle c\) shows \(\displaystyle x^{n}\), and hence every polynomial, is continuous on \(\displaystyle \mathbf{R}\).)

The rule changes only at \(\displaystyle x=1\): \(\displaystyle f(x)=x\) for \(\displaystyle x\le 1\) and \(\displaystyle f(x)=5\) for \(\displaystyle x>1\).At \(\displaystyle x=0\): all points near \(\displaystyle 0\) satisfy \(\displaystyle x\le 1\), so \(\displaystyle f(x)=x\) there. Hence \(\displaystyle \lim_{x\to 0}f(x)=0\) and \(\displaystyle f(0)=0\). They are equal, so \(\displaystyle f\) is continuous at \(\displaystyle x=0\).At \(\displaystyle x=1\): the two one-sided limits must be computed from different rules. \[\lim_{x\to 1^{-}}f(x)=\lim_{x\to 1^{-}}x=1,\qquad \lim_{x\to 1^{+}}f(x)=\lim_{x\to 1^{+}}5=5.\] Since LHL \(\displaystyle \neq\) RHL, \(\displaystyle \lim_{x\to 1}f(x)\) does not exist, so \(\displaystyle f\) is discontinuous at \(\displaystyle x=1\).At \(\displaystyle x=2\): all points near \(\displaystyle 2\) satisfy \(\displaystyle x>1\), so \(\displaystyle f(x)=5\) there. Hence \(\displaystyle \lim_{x\to 2}f(x)=5=f(2)\), and \(\displaystyle f\) is continuous at \(\displaystyle x=2\).Answer: \(\displaystyle f\) is continuous at \(\displaystyle x=0\) and at \(\displaystyle x=2\), and discontinuous at \(\displaystyle x=1\).

\(\displaystyle f(x)=2x+3\) for \(\displaystyle x\le 2\) and \(\displaystyle f(x)=2x-3\) for \(\displaystyle x>2\). On each of \(\displaystyle (-\infty,2)\) and \(\displaystyle (2,\infty)\) the function is a polynomial, hence continuous; only the joining point \(\displaystyle x=2\) needs testing.At \(\displaystyle x=2\): \[\lim_{x\to 2^{-}}f(x)=\lim_{x\to 2^{-}}(2x+3)=7,\qquad \lim_{x\to 2^{+}}f(x)=\lim_{x\to 2^{+}}(2x-3)=1,\] and \(\displaystyle f(2)=2(2)+3=7\).Since LHL \(\displaystyle =7\neq 1=\) RHL, the limit at \(\displaystyle x=2\) does not exist.Answer: \(\displaystyle x=2\) is the only point of discontinuity of \(\displaystyle f\).

For \(\displaystyle x\le -3\), \(\displaystyle |x|=-x\), so \(\displaystyle f(x)=-x+3\) there. Thus \[f(x)=\begin{cases}-x+3,&x\le -3\\ -2x,&-3<x<3\\ 6x+2,&x\ge 3.\end{cases}\] Each piece is a polynomial, so \(\displaystyle f\) is continuous at every point of \(\displaystyle (-\infty,-3)\), \(\displaystyle (-3,3)\) and \(\displaystyle (3,\infty)\). Only \(\displaystyle x=-3\) and \(\displaystyle x=3\) need testing.At \(\displaystyle x=-3\): \[\lim_{x\to -3^{-}}(-x+3)=3+3=6,\qquad \lim_{x\to -3^{+}}(-2x)=6,\qquad f(-3)=|-3|+3=6.\] All three agree, so \(\displaystyle f\) is continuous at \(\displaystyle x=-3\).At \(\displaystyle x=3\): \[\lim_{x\to 3^{-}}(-2x)=-6,\qquad \lim_{x\to 3^{+}}(6x+2)=20,\qquad f(3)=6(3)+2=20.\] Since LHL \(\displaystyle =-6\neq 20=\) RHL, the limit does not exist at \(\displaystyle x=3\).Answer: \(\displaystyle x=3\) is the only point of discontinuity.

Remove the modulus: for \(\displaystyle x>0\), \(\displaystyle |x|=x\) so \(\displaystyle \dfrac{|x|}{x}=1\); for \(\displaystyle x<0\), \(\displaystyle |x|=-x\) so \(\displaystyle \dfrac{|x|}{x}=-1\). Hence \[f(x)=\begin{cases}-1,&x<0\\ 0,&x=0\\ 1,&x>0.\end{cases}\] On \(\displaystyle (-\infty,0)\) and on \(\displaystyle (0,\infty)\), \(\displaystyle f\) is constant, hence continuous. Only \(\displaystyle x=0\) needs testing.At \(\displaystyle x=0\): \[\lim_{x\to 0^{-}}f(x)=-1,\qquad \lim_{x\to 0^{+}}f(x)=1,\qquad f(0)=0.\] Since LHL \(\displaystyle \neq\) RHL, \(\displaystyle \lim_{x\to 0}f(x)\) does not exist.Answer: \(\displaystyle x=0\) is the only point of discontinuity of \(\displaystyle f\).

Simplify the first branch. For \(\displaystyle x<0\) we have \(\displaystyle |x|=-x\), so \[\frac{x}{|x|}=\frac{x}{-x}=-1.\] Therefore \(\displaystyle f(x)=-1\) for \(\displaystyle x<0\) and \(\displaystyle f(x)=-1\) for \(\displaystyle x\ge 0\); that is, \(\displaystyle f(x)=-1\) for every real \(\displaystyle x\).A constant function is continuous: for any real \(\displaystyle c\), \(\displaystyle \lim_{x\to c}f(x)=-1=f(c)\). In particular at the joining point \(\displaystyle x=0\), LHL \(\displaystyle =\) RHL \(\displaystyle =f(0)=-1\).Answer: \(\displaystyle f\) is continuous on \(\displaystyle \mathbf{R}\); it has no point of discontinuity.

\(\displaystyle f(x)=x^{2}+1\) for \(\displaystyle x<1\) and \(\displaystyle f(x)=x+1\) for \(\displaystyle x\ge 1\). Each piece is a polynomial, so \(\displaystyle f\) is continuous on \(\displaystyle (-\infty,1)\) and on \(\displaystyle (1,\infty)\); only \(\displaystyle x=1\) needs testing.At \(\displaystyle x=1\): \[\lim_{x\to 1^{-}}f(x)=\lim_{x\to 1^{-}}\left(x^{2}+1\right)=1+1=2,\qquad \lim_{x\to 1^{+}}f(x)=\lim_{x\to 1^{+}}(x+1)=2,\] and \(\displaystyle f(1)=1+1=2\).Since LHL \(\displaystyle =\) RHL \(\displaystyle =f(1)=2\), \(\displaystyle f\) is continuous at \(\displaystyle x=1\) as well.Answer: \(\displaystyle f\) is continuous on \(\displaystyle \mathbf{R}\); there is no point of discontinuity.