Question

Find the limit of the function
$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\cos 2x-1}{\cos x-1}$

Answer 1

Hint : In this question, we first need to convert the given numerator using the trigonometric ratios of multiple angles given by $\cos 2A={{\cos }^{2}}A-{{\sin }^{2}}A$ and then use the trigonometric identity given by ${{\cos }^{2}}A+{{\sin }^{2}}A=1$ and simplify further. Now, expand the numerator using the factorisation of polynomials given by $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ and cancel the common terms in numerator and denominator and then apply the given limit to simplify further and get the result.

Complete step-by-step answer :
TRIGONOMETRIC IDENTITY:
An equation involving trigonometric functions which is true for all those angles for which the functions are defined is called trigonometric identity.
Now, one of the trigonometric identity is given by
${{\cos }^{2}}A+{{\sin }^{2}}A=1$
As we already know from the trigonometric ratios of compound angles that
$\cos 2A={{\cos }^{2}}A-{{\sin }^{2}}A$
Now, from the factorisation of polynomials we also have that
$\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$
Now, from the given limit in the question we have
$\Rightarrow \underset{x\to 0}{\mathop{\lim }}\,\dfrac{\cos 2x-1}{\cos x-1}$
Now, let us rewrite the numerator above using the trigonometric ratios of compound angles
$\cos 2A={{\cos }^{2}}A-{{\sin }^{2}}A$
Now, on using the above formula we can write it as
$\Rightarrow \underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{\cos }^{2}}x-{{\sin }^{2}}x-1}{\cos x-1}$
Now, further rewrite it in the simple form using the trigonometric identity
${{\sin }^{2}}A=1-{{\cos }^{2}}A$
Now, on further substituting this identity we get,
$\Rightarrow \underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{\cos }^{2}}x-\left( 1-{{\cos }^{2}}x \right)-1}{\cos x-1}$
Now, this can be further also written as
$\Rightarrow \underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{\cos }^{2}}x+{{\cos }^{2}}x-1-1}{\cos x-1}$
Now, on further writing it in the simplified form we get,
$\Rightarrow \underset{x\to 0}{\mathop{\lim }}\,\dfrac{2\left( {{\cos }^{2}}x-1 \right)}{\cos x-1}$
Let us now further factorise the numerator using the factorisation of polynomials given by
$\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$
Now, on using the above factorisation we get,
$\Rightarrow \underset{x\to 0}{\mathop{\lim }}\,\dfrac{2\left( \cos x-1 \right)\left( \cos x+1 \right)}{\cos x-1}$
Now, on cancelling the common term in the numerator and denominator we get,
$\Rightarrow \underset{x\to 0}{\mathop{\lim }}\,2\left( \cos x+1 \right)$
Now, this can be further written as
$\Rightarrow 2\underset{x\to 0}{\mathop{\lim }}\,\left( \cos x+1 \right)$
Now, on further applying the limit we get,
$\Rightarrow 2\left( \cos {{0}^{\circ }}+1 \right)$
Now, this can be further written as
$\Rightarrow 2\left( 1+1 \right)\text{ }\left[ \because \cos {{0}^{\circ }}=1 \right]$
Now, on further simplification we get,
$\Rightarrow 4\text{ }$
Hence, the value of the limit is 4.

Note : Instead of using the compound angle and trigonometric identity and the factorisation we can also solve it by converting the numerator using the compound angle formula into sine term and the denominator using half angle formula and the simplify them using the properties of limits which gives the same result.It is important to note that we need to use the respective formulae to rewrite the numerator and to simplify further because if we consider incorrect sign or neglect any term then the result would be incorrect.