Question

Find the value of $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{x}^{3}}\cot x}{1-\cos x}$.

Answer 1

Hint: Convert $\cot x$ as $\left( \dfrac{\cos x}{\sin x} \right)$. Use identities such as $\left\{ \underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{x}=1 \right\}$ and also L’Hopital’s rule.

Complete step-by-step answer:

In the question we have to find limit,

$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{x}^{3}}\cot x}{1-\cos x}$

So, now we will put $\cot x=\left( \dfrac{\cos x}{\sin x} \right)$, so the above equation can be written as,

$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{x}^{3}}\cot x}{1-\cos x}=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{x}^{3}}\cos x}{\sin x(1-\cos x)}$

$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{x}^{3}}\cot x}{1-\cos x}=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{x}^{2}}\cos x}{\dfrac{sinx}{x}(1-\cos x)}$

Here , $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{x}=1$. So this identity we can use here, so the above expression can be written as,

$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{x}^{3}}\cot x}{1-\cos x}=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{x}^{2}}\cos x}{(1-\cos x)}$

Now divide $\cos x$ from denominator and numerator, we get

$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{x}^{3}}\cot x}{1-\cos x}=\underset{x\to

0}{\mathop{\lim }}\,\dfrac{{{x}^{2}}}{(\sec x-1)}$

If we put x=0, then ,

$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{x}^{3}}\cot x}{1-\cos x}=\dfrac{{{x}^{2}}}{(\sec

x-1)}=\dfrac{0}{\sec 0-1}$

As we know that the value of $\sec 0=1$, So,

$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{x}^{3}}\cot x}{1-\cos

x}=\dfrac{0}{1-1}=\dfrac{0}{0}$

So, $\dfrac{{{x}^{2}}}{(\sec x-1)}$ is of ‘$\dfrac{0}{0}$’ form.

When a limit value is of ‘$\dfrac{0}{0}$’ form then we use L’Hopital’s rule.

L’Hopital’s rule evaluates by differentiating the numerator and denominator independently until a determinant form comes.

So, applying L’Hopital’s rule, we get,

$\begin{align}

  & \underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{x}^{3}}\cot x}{1-\cos x}=\dfrac{\dfrac{d}{dx}\left( {{x}^{2}} \right)}{\dfrac{d}{dx}(\sec x-1)} \\

 & \Rightarrow \underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{x}^{3}}\cot x}{1-\cos x}=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{2x}{\sec x\tan x} \\

\end{align}$

Dividing ‘x’ from both numerator and denominator we get,

$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{x}^{3}}\cot x}{1-\cos x}=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{2}{\sec x\dfrac{\tan x}{x}}$

As we know $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\tan x}{x}=1$. So this identity we can use here.

$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{x}^{3}}\cot x}{1-\cos x}=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{2}{\sec x\times 1}=\underset{x\to 0}{\mathop{\lim }}\,2\cos x$
We know that $\cos 0=1$, so applying the limits, we get

$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{x}^{3}}\cot x}{1-\cos x}=2\times \cos 0=2\times 1=2$

So, the value of the given limit is ‘2’.


Note: In these types of questions students should be careful while calculations because a single mistake can make the whole question wrong.
Another approach is we can directly apply limits in $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{x}^{3}}\cot x}{1-\cos x}$, we get $\dfrac{0}{0}$ form then we use L’Hopital’s rule, we will get the same answer.