Question:
Let $f(x)=\left\{\begin{array}{l}4 x-5, x \leq 2 \\ x-a, x>2\end{array}\right.$
If $\lim _{x \rightarrow 2} f(x)$ exists then find the value of $a$.
Solution:
Left Hand Limit(L.H.L.):
$\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{-}} 4 x-5$
$=4(2)-5$
$=8-5$
$=3$
Right Hand Limit(R.H.L.):
$\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{+}} x-a$
$=2-\mathrm{a}$
Since $\lim _{x \rightarrow 2} f(x)$ it exists,
$\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{+}} f(x)$
$\rightarrow 3=2-a$
$\rightarrow a=2-3$
$\rightarrow a=-1$
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