Question

How do you find the limit of ${{\left( \cos x \right)}^{\dfrac{1}{{{x}^{2}}}}}$ as $x \to 0$?

Answer 1

Hint: We first need to take the limit form of the function ${{\left( \cos x \right)}^{\dfrac{1}{{{x}^{2}}}}}$. Then for the power reduction we use logarithm on both sides. Applying the limit, we get an indeterminate form where we take L’hopital rule to get the differentiated form. Then we apply the limit to get the value of \[\log p\]. We take exponential form to get the value of the limit.

Complete step-by-step solution:
We need to find the value of $\displaystyle \lim_{x \to 0}{{\left( \cos x \right)}^{\dfrac{1}{{{x}^{2}}}}}$.
We assume $p=\displaystyle \lim_{x \to 0}{{\left( \cos x \right)}^{\dfrac{1}{{{x}^{2}}}}}$ and take logarithm both sides.
\[\log p=\log \left[ \displaystyle \lim_{x \to 0}{{\left( \cos x \right)}^{\dfrac{1}{{{x}^{2}}}}} \right]=\displaystyle \lim_{x \to 0}\left[ \log {{\left( \cos x \right)}^{\dfrac{1}{{{x}^{2}}}}} \right]\].
We know that $\log {{a}^{b}}=b\log a$.
Therefore, \[\log p=\displaystyle \lim_{x \to 0}\left[ \dfrac{\log \left( \cos x \right)}{{{x}^{2}}} \right]\].
Now when we are applying the limit value of the variable x in the function \[\dfrac{\log \left( \cos x \right)}{{{x}^{2}}}\], we get the value of \[\dfrac{\log \left( \cos x \right)}{{{x}^{2}}}=\dfrac{\log \left( \cos 0 \right)}{{{0}^{2}}}=\dfrac{0}{0}\]. This becomes an indeterminate form.
We have to apply the L’hopital rule.
In that case we take the differentiated form of the functions separately in the denominator and the numerator. Then we apply the limit value again.
If we get a value for the limit then that becomes the answer and if we get an indeterminant form again, we continue the process.
Applying the L’hopital rule on the function \[\dfrac{\log \left( \cos x \right)}{{{x}^{2}}}\], we get \[\dfrac{\dfrac{d}{dx}\left[ \log \left( \cos x \right) \right]}{\dfrac{d}{dx}\left[ {{x}^{2}} \right]}=\dfrac{\dfrac{-\sin x}{\cos x}}{2x}=\dfrac{-\tan x}{2x}\].
We apply the limit value of $x \to 0$ and get \[\dfrac{-\tan x}{2x}=\dfrac{-\tan 0}{2\times 0}=\dfrac{0}{0}\]. This becomes an indeterminate form again.
We have to apply the L’hopital rule.
\[\dfrac{\dfrac{d}{dx}\left[ -\tan x \right]}{\dfrac{d}{dx}\left[ 2x \right]}=\dfrac{-{{\sec }^{2}}x}{2}\].
Now we apply the limit and get \[\log p=\displaystyle \lim_{x \to 0}\left[ \dfrac{\log \left( \cos x \right)}{{{x}^{2}}} \right]=\displaystyle \lim_{x \to 0}\left[ \dfrac{-{{\sec }^{2}}x}{2} \right]\].
\[\log p=\displaystyle \lim_{x \to 0}\left[ \dfrac{-{{\sec }^{2}}x}{2} \right]=\dfrac{-{{\sec }^{2}}0}{2}=\dfrac{-1}{2}\].
We know that ${{\log }_{x}}a=b\Rightarrow {{x}^{b}}=a$.
Now we need to find the value of p and that’s why we take exponential form and get
\[\begin{align}
  & {{\log }_{e}}p=\dfrac{-1}{2} \\
 & \Rightarrow p={{e}^{\dfrac{-1}{2}}}=\dfrac{1}{\sqrt{e}} \\
\end{align}\]
Therefore, the value of $\displaystyle \lim_{x \to 0}{{\left( \cos x \right)}^{\dfrac{1}{{{x}^{2}}}}}$ is \[\dfrac{1}{\sqrt{e}}\].

Note: The indeterminate form of the limit is achieved when we have $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$ form. This is called indeterminate form as the behaviour of the function \[\displaystyle \lim_{x \to 0}\left[ \dfrac{g\left( x \right)}{f\left( x \right)} \right]\] it can’t be determined. So, we always use the L’hopital rule to tackle the problem.