Question

What is the limit $\displaystyle \lim_{x \to 0}\dfrac{\cos x-1}{x}$?

Answer 1

Hint: Assume the given limit as L. Use the trigonometric identity $\cos x=1-2si{{n}^{2}}\left( \dfrac{x}{2} \right)$ and simplify the numerator. Now, try to make the denominator equal to the argument of the sine function. For this, multiply the denominator with $\dfrac{x}{4}$ and to balance this change multiply the numerator also with the same. Use the formula $\displaystyle \lim_{\theta \to 0}\dfrac{\sin \theta }{\theta }=1$ to get the answer.

Complete step by step solution:
Here we have been provided with the expression $\displaystyle \lim_{x \to 0}\dfrac{\cos x-1}{x}$ and we are asked to find its value. Let us assume this limit as L, so we have,
$\Rightarrow L=\displaystyle \lim_{x \to 0}\dfrac{\cos x-1}{x}$
Here we can see that we cannot directly substitute the value of x to get the answer because if we will do so then the limit will become $\dfrac{0}{0}$ which is indeterminate form. So we need a different approach. Now, using the trigonometric identity $\cos x=1-2si{{n}^{2}}\left( \dfrac{x}{2} \right)$ we get,
$\begin{align}
 & \Rightarrow L=\displaystyle \lim_{x \to 0}\dfrac{1-2{{\sin }^{2}}\left( \dfrac{x}{2} \right)-1}{x} \\
 & \Rightarrow L=\displaystyle \lim_{x \to 0}\dfrac{-2{{\sin }^{2}}\left( \dfrac{x}{2} \right)}{x} \\
 & \Rightarrow L=\displaystyle \lim_{x \to 0}\dfrac{-2{{\left[ \sin \left( \dfrac{x}{2} \right) \right]}^{2}}}{x} \\
\end{align}$
Multiplying the numerator and denominator with $\dfrac{x}{4}$ we get,
\[\begin{align}
 & \Rightarrow L=\displaystyle \lim_{x \to 0}\dfrac{-2{{\left[ \sin \left( \dfrac{x}{2} \right) \right]}^{2}}}{x\times \dfrac{x}{4}}\times \dfrac{x}{4} \\
 & \Rightarrow L=\displaystyle \lim_{x \to 0}\dfrac{-2{{\left[ \sin \left( \dfrac{x}{2} \right) \right]}^{2}}}{{{\left( \dfrac{x}{2} \right)}^{2}}}\times \dfrac{x}{4} \\
 & \Rightarrow L=\displaystyle \lim_{x \to 0}\left( -2 \right)\times {{\left[ \dfrac{\sin \left( \dfrac{x}{2} \right)}{\left( \dfrac{x}{2} \right)} \right]}^{2}}\times \dfrac{x}{4} \\
\end{align}\]
The expression can also be written as:
\[\Rightarrow L=\left( -2 \right)\times {{\left[ \displaystyle \lim_{x \to 0}\dfrac{\sin \left( \dfrac{x}{2} \right)}{\left( \dfrac{x}{2} \right)} \right]}^{2}}\times \displaystyle \lim_{x \to 0}\dfrac{x}{4}\]
We know that $\displaystyle \lim_{\theta \to 0}\dfrac{\sin \theta }{\theta }=1$, so we get,
\[\begin{align}
 & \Rightarrow L=\left( -2 \right)\times {{\left[ 1 \right]}^{2}}\times \dfrac{0}{4} \\
 & \therefore L=0 \\
\end{align}\]

Hence, the value of the given expression of limit is 0.

Note: There are two other methods by which you can solve the above question. The easiest one is called the L Hospital’s rule. This method is used when we have the limit of the form $\displaystyle \lim_{x \to 0}\dfrac{f\left( x \right)}{g\left( x \right)}=\dfrac{0}{0}$ or $\displaystyle \lim_{x \to 0}\dfrac{f\left( x \right)}{g\left( x \right)}=\dfrac{\infty }{\infty }$. Here what we do is, we differentiate $f\left( x \right)$ and $g\left( x \right)$ to get the limit of the form $\displaystyle \lim_{x \to 0}\dfrac{f'\left( x \right)}{g'\left( x \right)}$ and check if still it is in indeterminate form or not. If it is not then we substitute the value of x to get the answer and if it is still in indeterminate form then we again differentiate and this process continues till it reaches its determinate form. The other method is the expansion formula of $\cos x$ given as $\cos x=1+\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{4}}}{4!}+\dfrac{{{x}^{6}}}{6!}+......$. Substitute this value in the limit and cancel the like terms and at last substitute the value of x to get the answer.