For what value of $\lambda$ is the function defined by $f(x) = \left\{ {\begin{array}{llllllllllllllllllll}{\lambda ({x^2} - 2x),}&{if}&{x \le 0}\\{4x + 1,}&{if}&{x > 0}\end{array}} \right.$ continuous at x$=$ 0?}}

What about continuity at x $=$ 1?

Since f (x) is continuous at x $=$ 0,

(i) $\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ - }} \lambda ({x^2} - 2x) = \lambda (0 - 0) = 0$

$\mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} 4x + 1 = 4(0) + 1 = 1$

As L.H.L. $\ne$ R.H.L.

Hence we can say that f(x) is continuous at x $=$ 0 for no value of $\lambda$.

(ii) At x $=$ 1 :
$\mathop {\lim }\limits_{x \to 1} f(x) = \mathop {\lim }\limits_{x \to 1} 4x + 1 = 4 \times 1 + 1 = 5$
and $f(1) = 4(1) + 1 = 5$

Thus,$\mathop {\lim }\limits_{x \to 1} f(x) = f(1)$ for any value of $\lambda$.

Hence we can say that f(x) is continuous at x $=$ 1 for any real value of $\lambda$ .