Prove the following
Question:

$f(x)=\left\{\begin{array}{ll}3 x-8, & \text { if } x \leq 5 \\ 2 k, & \text { if } x>5\end{array}\right.$ at $x=5$

Solution:

Finding the left hand and right hand limits for the given function, we have

$\lim _{x \rightarrow 5} f(x)=3 x-8$

$=\lim _{h \rightarrow 0} 3(5-h)-8=15-8=7$

$\lim _{x \rightarrow 5^{+}} f(x)=2 k$

As the function is continuous at $x=5$

$\lim _{x \rightarrow 5^{-}} f(x)=\lim _{x \rightarrow 5^{+}} f(x)$

So,

7 = 2k

k = 7/2 = 3.5

Therefore, the value of k is 3.5

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