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Find the value of k, if
$f\left( x \right)=\dfrac{1-\cos 2x}{1-\cos x}$, $x\ne 0$
$f\left( x \right)=k$ , $x=0$

Last updated date: 14th Jul 2024
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Hint: Consider a curve defined by a function $y=f\left( x \right)$ passing through the vertical axis. At a point, $x=0$ the curve cuts the axis. If we substitute the value $x=0$ in the given function $y=f\left( x \right)$, we will get the output y at which the curve cuts the vertical axis. This is the general procedure for finding the coordinates of the particular point.

Complete Step by Step Solution:
It is given that when the variable x is not equal to zero, the function $f\left( x \right)$ is equal to
$\Rightarrow f\left( x \right)=\dfrac{1-\cos 2x}{1-\cos x}$
The above equation cannot give the output value at $x=0$ because the value of $\cos {{0}^{\circ }}$ is equal to 1.
$\Rightarrow f\left( x \right)=\dfrac{1-\cos 2x}{1-\cos {{0}^{\circ }}}=\dfrac{1-\cos 2x}{1-1}=\infty $
We can get very close to zero $\left( x\to 0 \right)$ on both sides of the axis to find the closest value we can arrive at. In the left-hand side limit (LHL), we are approaching from a negative value to zero $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,$. Similarly, we are approaching from the positive side to zero $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,$in the right-hand side limit (RHL).
Let us assume that the left hand and the right-hand limits are equal at $x=0$.
$\Rightarrow LHL=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)=k$
$\Rightarrow RHL=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{1-\cos 2x}{1-\cos x}$
We know the trigonometric identity $1-\cos 2x=2{{\sin }^{2}}x$ . Substitute them in the above equation.
$\Rightarrow RHL=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{2{{\sin }^{2}}x}{1-\cos x}$
From the trigonometric identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$
$\Rightarrow RHL=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{2\left( 1-{{\cos }^{2}}x \right)}{1-\cos x}=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{2\left( {{1}^{2}}-{{\cos }^{2}}x \right)}{1-\cos x}$
We know that $\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a+b \right)\left( a-b \right)$
$\Rightarrow RHL=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{2\left( 1+\cos x \right)\left( 1-\cos x \right)}{1-\cos x}=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,2\left( 1+\cos x \right)$
The given function is continuous, therefore at the point $x=0$
$\Rightarrow RHL=2\left( 1+\cos {{0}^{\circ }} \right)=2\left( 1+1 \right)$
$\Rightarrow RHL=4$
Equating the left-hand and right-hand limits, we get
$\Rightarrow k=4$

Sometimes, the function may be non-continuous and it will break at the point where it cuts the axis. In other words, the curve will break at the point where it meets the axis and it will be away from the point where the curve starts to progress on the other side of the axis. Therefore the function will give two different outputs when given an input equal to zero.