Question

f(x)= \[\left[ \begin{align}
  & \dfrac{\sin x-\cos x}{x-\dfrac{\pi }{4}}\ \ x\ne \dfrac{\pi }{4} \\
 & k\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x=\dfrac{\pi }{4} \\
\end{align} \right]\] if function f is continuous at x= \[\dfrac{\pi }{4}\] , find k.

Answer 1

Hint: If a function is continuous at a particular value of x, then this implies that the left hand limit that is L.H.L and the right hand limit that is R.H.L exist and they are equal to the value of the function at that particular value of x.
The formula that will be used in the solution is as follows:
If $f\left( x \right)=\dfrac{g\left( x \right)}{h\left( x \right)}$ and $\underset{x\to t}{\mathop{\lim }}\,\dfrac{g\left( x \right)}{h\left( x \right)}=\dfrac{0}{0}$ or $\underset{x\to t}{\mathop{\lim }}\,\dfrac{g\left( x \right)}{h\left( x \right)}=\dfrac{\infty }{\infty }$ then the value of limit is given as\[\underset{x\to t}{\mathop{\lim }}\,f(x)=\underset{x\to t}{\mathop{\lim }}\,\dfrac{g(x)}{h(x)}=\dfrac{{g}'(t)}{{h}'(t)}\] .
The above formula is known as L’Hospital rule and it is used to find the limit of a particular function.

Complete step-by-step answer:
As mentioned in the question, we have to find the value of k by checking the continuity of the given function.
We will first find the value of LHL.
Firstly, we will put the limit in the function and then see what value is obtained.
Now, LHL is given as:
\[\begin{align}
  & LHL=\underset{x\to {{\dfrac{\pi }{4}}^{-}}}{\mathop{\lim }}\,f(x) \\
 & \Rightarrow LHL=\underset{x\to {{\dfrac{\pi }{4}}^{-}}}{\mathop{\lim }}\,\left( \dfrac{\sin x-\cos x}{x-\dfrac{\pi }{4}}\ \right) \\
\end{align}\]
Putting the value in the function we get:
\[\begin{align}
  & LHL=\underset{x\to {{\dfrac{\pi }{4}}^{-}}}{\mathop{\lim }}\,\left( \dfrac{\sin x-\cos x}{x-\dfrac{\pi }{4}}\ \right) \\
 &\Rightarrow LHL=\left( \dfrac{\sin {{\dfrac{\pi }{4}}^{-}}-\cos {{\dfrac{\pi }{4}}^{-}}}{{{\dfrac{\pi }{4}}^{-}}-\dfrac{\pi }{4}} \right) \\
 &\Rightarrow LHL=\left( \dfrac{\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{2}}}{0} \right) \\
 &\Rightarrow LHL=\dfrac{0}{0} \\
\end{align}\]
The LHL comes out as $\dfrac{0}{0}$.
Now, we know that when a limit comes out in the form of $\dfrac{0}{0}$, we can solve it by using L-Hospital rule given as:
$\underset{x\to t}{\mathop{\lim }}\,\dfrac{g\left( x \right)}{h\left( x \right)}=\underset{x\to t}{\mathop{\lim }}\,\dfrac{g'\left( x \right)}{h'\left( x \right)}$.
Using L-H rule for obtaining LHL, we get:
\[LHL=\underset{x\to {{\dfrac{\pi }{4}}^{-}}}{\mathop{\lim }}\,\left( \dfrac{\sin x-\cos x}{x-\dfrac{\pi }{4}}\ \right)\]
Differentiating the numerator and the denominator w.r.t. x, we get:
$\begin{align}
  & LHL=\underset{x\to {{\dfrac{\pi }{4}}^{-}}}{\mathop{\lim }}\,\left( \dfrac{\cos x-\left( -\sin x \right)}{1-0} \right) \\
 &\Rightarrow LHL=\underset{x\to {{\dfrac{\pi }{4}}^{-}}}{\mathop{\lim }}\,\left( \cos x+\sin x \right) \\
\end{align}$
Putting the value of the limit, we get:
$\begin{align}
  & LHL=\underset{x\to {{\dfrac{\pi }{4}}^{-}}}{\mathop{\lim }}\,\left( \sin x+\cos x \right) \\
 &\Rightarrow LHL=\sin {{\dfrac{\pi }{4}}^{-}}+\cos {{\dfrac{\pi }{4}}^{-}} \\
 &\Rightarrow LHL=\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}} \\
 &\Rightarrow LHL=\dfrac{2}{\sqrt{2}} \\
 &\Rightarrow LHL=\sqrt{2} \\
\end{align}$
Thus, the value of LHL is $\sqrt{2}$.
Similarly,
For finding the RHL, we write the expression as
\[\begin{align}
  & RHL=\underset{x\to {{\dfrac{\pi }{4}}^{+}}}{\mathop{\lim }}\,f(x) \\
 &\Rightarrow RHL=\underset{x\to {{\dfrac{\pi }{4}}^{+}}}{\mathop{\lim }}\,\left( \dfrac{\sin x-\cos x}{x-\dfrac{\pi }{4}}\ \right) \\
 &\Rightarrow RHL=\dfrac{\sin {{\dfrac{\pi }{4}}^{+}}-\cos {{\dfrac{\pi }{4}}^{+}}}{{{\dfrac{\pi }{4}}^{+}}-\dfrac{\pi }{4}} \\
 &\Rightarrow RHL=\dfrac{\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{2}}}{0} \\
 &\Rightarrow RHL=\dfrac{0}{0} \\
\end{align}\]
Thus, we will use L-H rule to find the RHL. Using L-H rule for finding out the RHL, we get:
\[RHL=\underset{x\to {{\dfrac{\pi }{4}}^{+}}}{\mathop{\lim }}\,\left( \dfrac{\sin x-\cos x}{x-\dfrac{\pi }{4}}\ \right)\]
Differentiating the numerator and the denominator w.r.t. x we get:
\[\begin{align}
  & RHL=\underset{x\to {{\dfrac{\pi }{4}}^{+}}}{\mathop{\lim }}\,\left( \dfrac{\sin x-\cos x}{x-\dfrac{\pi }{4}}\ \right) \\
 &\Rightarrow RHL=\underset{x\to {{\dfrac{\pi }{4}}^{+}}}{\mathop{\lim }}\,\left( \dfrac{\cos x-\left( -\sin x \right)}{1-0} \right) \\
 &\Rightarrow RHL=\underset{x\to {{\dfrac{\pi }{4}}^{+}}}{\mathop{\lim }}\,\left( \cos x+\sin x \right) \\
\end{align}\]
Putting the value of limit we get:
$\begin{align}
  & RHL=\underset{x\to {{\dfrac{\pi }{4}}^{+}}}{\mathop{\lim }}\,\left( \cos x+\sin x \right) \\
 &\Rightarrow RHL=\cos {{\dfrac{\pi }{4}}^{+}}+\sin {{\dfrac{\pi }{4}}^{+}} \\
 &\Rightarrow RHL=\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}} \\
 &\Rightarrow RHL=\dfrac{2}{\sqrt{2}}=\sqrt{2} \\
\end{align}$
Thus, the value of RHL is $\sqrt{2}$.
Now, as L.H.L and R.H.L exist and are equal, hence, the value of the function is equal to this value as the function is continuous. Therefore, we get
\[\begin{align}
  & f\left( \dfrac{\pi }{4} \right)=k \\
 &\Rightarrow k=\sqrt{2} \\
\end{align}\]
Hence, the value of k is \[\sqrt{2}\].

Note: The students can make an error if finding the value of k if they don’t know the basic definition of continuity of a function which is as follows:
If a function is continuous at a particular value of x, then this implies that the left hand limit that is L.H.L and the right hand limit that is R.H.L exist and they are equal to the value of the function at that particular value of x.
Also, we have here used the L-H rule. Remember that this rule is only applied when after keeping the value of limit in the function, we get $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$.