Question

Find the value of k so that the function f(x) = $\left\{ \begin{align}
  & kx+1,\,\,\,if\,x\le 5 \\
 & 3x-5,\,\,if\,x>5 \\
\end{align} \right.$ at x = 5 is a continuous function.

Answer 1

Hint: To solve this problem we need to know that for a function f(x) to be continuous at x = ‘a’ then for that limit of the function should exist at that point i.e. left-hand limit and right-hand limit both should be equal at x = ‘a’ and also the obtained limit should be equal to the value of that function at the point ‘a’. So to solve this we will first find the right-hand limit of the function at x = 5 and then equate that obtained limit with the value of the left-hand limit of the function at x = 5. And from that, we will get our answer.

Complete step by step answer:
To solve this we have to know that for a function to be continuous at a given point then its limit should exist at that point and that limit should be equal to the value of the function at that point.
So first we will calculate the limit of the function at the given point i.e. x = 5,
So right hand limit of the function at x = 5 is given as,
$\begin{align}
  & \displaystyle \lim_{x \to {{5}^{-}}}f\left( x \right) \\
 & =\displaystyle \lim_{h\to 0}f\left( 5-h \right) \\
 & =\displaystyle \lim_{h\to 0}\left( k\left( 5-h \right)+1 \right) \\
 & =k\left( 5-0 \right)+1 \\
 & =5k+1 \\
\end{align}$
And the left hand limit is given as,
$\begin{align}
  & \displaystyle \lim_{x \to {{5}^{+}}}f\left( x \right) \\
 & =\displaystyle \lim_{h\to 0}f\left( 5+h \right) \\
 & =\displaystyle \lim_{h\to 0}\left( 3\left( 5+h \right)-5 \right) \\
 & =3\left( 5+0 \right)-5 \\
 & =15-5 \\
 & =10 \\
\end{align}$
Now we know that for limit to exist LHL = RHL, so we get
\[\begin{align}
  & 5k+1=10 \\
 & \Rightarrow 5k=10-1 \\
 & \Rightarrow 5k=9 \\
 & \Rightarrow k=\dfrac{9}{5} \\
\end{align}\]
Hence we get value of k as $\dfrac{9}{5}$

Note:
 You should note and remember that if the function is continuous at a point ‘a’ then its limit must exist at that point whereas the converse is not true i.e. if the limit exists at point ‘a’ it may or may not be continuous at that point. Also in the above question remember to equate the Left-hand limit with the right-hand limit some students do not equate and directly write the answer as RHL but it is wrong you have to equate it with the value to get the answer.