$f(x) = \left\{ {\begin{array}{llllllllllllllllllll}{{x^{10}} - 1,}&{ifx \le 1}\\{{x^2},}&{ifx > 1}\end{array}} \right.$

Weobserve thatf(x)is continuousat real numbers x < 1 and x > 1 as it is polynomial function.
Now, continuity at x $=$ 1:

L.H.L.$=$ $\mathop {\lim }\limits_{x \to {1^ - }} f(x) = \mathop {\lim }\limits_{x \to {1^ - }} ({x^{10}} - 1) = \mathop {\lim }\limits_{\scriptstyle x \to 1 - h\atop\scriptstyle h \to 0} [{(1 - h)^{10}} - 1] = 0$

R.H.L.$= \mathop {\lim }\limits_{x \to {1^ + }} f(x) = \mathop {\lim }\limits_{x \to {1^ + }} ({x^2}) = \mathop {\lim }\limits_{\scriptstyle x \to 1 + h\atop\scriptstyle h \to 0} {(1 + h)^2} = 1$

Also, $f(1) = {1^{10}} - 1 = 0$
therefore, L.H.L.$\ne$ R.H.L.$\ne f(1)$

Hence we can say that f is discontinuous at x $=$ 1.

So, the only point of discontinuity of f (x) is 1.