Question

In each of the following, find the value of the constant k so that the given function is continuous at the indicated point;  

\[f\left( x \right) = \begin{cases}k( x^2 - 2x), \text{ if }  & x < 0 \\ \cos x, \text{ if }  & x \geq 0\end{cases}\] at x = 0

Answer 1

Given:

\[f\left( x \right) = \binom{k\left( x^2 - 2x \right), \text{ if } x < 0}{\ cosx, \text{ if }  x \geq 0}\] 

We have

(LHL at x = 0) = 

\[\lim_{x \to 0^-} f\left( x \right) = \lim_{h \to 0} f\left( 0 - h \right) = \lim_{h \to 0} f\left( - h \right) = \lim_{h \to 0} k\left( h^2 + 2h \right) = 0\]

\[\lim_{x \to 0^+} f\left( x \right) = \lim_{h \to 0} f\left( 0 + h \right) = \lim_{h \to 0} f\left( h \right) = \lim_{h \to 0} \cosh = 1\]

\[\therefore \lim_{x \to 0^-} f\left( x \right) \neq \lim_{x \to 0^+} f\left( x \right)\]

Thus, no value of k exists for which f\left( x \right) is continuous at x = 0 .