Question:
Let $f(x)=\left\{\begin{array}{l}\cos x, x \geq 0 \\ x+k, x<0\end{array}\right.$
Find the value of $k$ for which $\lim _{x \rightarrow 0} f(x)$ exist.
Solution:
Left Hand Limit(L.H.L.):
$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} x+k$
$=0+k$
$=k$
Right Hand Limit(R.H.L.):
$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}} \cos x$
$=\cos (0)$
$=1$
It is given that $\lim _{x \rightarrow 0} f(x)$ exists. Therefore,
$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)$
$\rightarrow \mathrm{k}=1$
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