Evaluate the limit $\underset{x\to \dfrac{\pi }{2}}{\mathop{\lim }}\,\left( \cos x\cdot \log \left( \tan x \right) \right)$.

Hint: We will use the method of substitution of the value that is in the limit. Also we will use L’Hopital’s Rule.

Complete step-by-step solution -
 Now, we will consider the limit as $\underset{x\to \dfrac{\pi }{2}}{\mathop{\lim }}\,\left( \cos x\cdot \log \left( \tan x \right) \right)....(i)$.
Now, we will substitute the value of the limit that is, we will put $x=\dfrac{\pi }{2}$ in expression (i). Therefore, we have $\underset{x\to \dfrac{\pi }{2}}{\mathop{\lim }}\,\left( \cos x\cdot \log \left( \tan x \right) \right)=\left( \cos \left( \dfrac{\pi }{2} \right)\cdot \log \left( \tan \left( \dfrac{\pi }{2} \right) \right) \right)$. As we know that value of $\cos \left( \dfrac{\pi }{2} \right)=0$ and the value of $\tan \left( \dfrac{\pi }{2} \right)=\infty $. Therefore, we have $\left( \cos \left( \dfrac{\pi }{2} \right)\cdot \log \left( \tan \left( \dfrac{\pi }{2} \right) \right) \right)=\left( 0\cdot \log \left( \infty \right) \right)$. Since, the value of $\log \left( \infty \right)=\infty $. So, we have that $\left( 0\cdot \log \left( \infty \right) \right)=0\cdot \infty $. Thus, it is a $0\cdot \infty $ case. So, we can apply L’Hopital’s Rule here. But before that we will convert the trigonometric term $\cos \left( x \right)=\dfrac{1}{\sec \left( x \right)}$ and apply this as a substitution to equation (i). Therefore, we have
$\begin{align}
  & \underset{x\to \dfrac{\pi }{2}}{\mathop{\lim }}\,\left( \cos x\cdot \log \left( \tan x \right) \right)=\underset{x\to \dfrac{\pi }{2}}{\mathop{\lim }}\,\left( \dfrac{1}{\sec \left( x \right)}\cdot \log \left( \tan x \right) \right) \\
 & \Rightarrow \underset{x\to \dfrac{\pi }{2}}{\mathop{\lim }}\,\left( \dfrac{1}{\sec \left( x \right)}\cdot \log \left( \tan x \right) \right)=\underset{x\to \dfrac{\pi }{2}}{\mathop{\lim }}\,\left( \dfrac{\log \left( \tan x \right)}{\sec \left( x \right)} \right) \\
\end{align}$
Now we will apply L’Hopital’s rule and differentiate numerator and denominator individually. As we know that the differentiation is $\log \left( \tan x \right)=\dfrac{1}{\tan \left( x \right)}\times {{\sec }^{2}}\left( x \right)$ and the differentiation of $\sec \left( x \right)=\sec \left( x \right)\tan \left( x \right)$. Thus, we have that $\underset{x\to \dfrac{\pi }{2}}{\mathop{\lim }}\,\left( \dfrac{\log \left( \tan x \right)}{\sec \left( x \right)} \right)=\underset{x\to \dfrac{\pi }{2}}{\mathop{\lim }}\,\left( \dfrac{\dfrac{{{\sec }^{2}}\left( x \right)}{\tan \left( x \right)}}{\dfrac{\sec \left( x \right)\tan \left( x \right)}{1}} \right)$. Now, we will apply the formula $\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}}=\dfrac{a}{b}\times \dfrac{d}{c}$ here at this step. Thus, we have that $\underset{x\to \dfrac{\pi }{2}}{\mathop{\lim }}\,\left( \dfrac{\dfrac{{{\sec }^{2}}\left( x \right)}{\tan \left( x \right)}}{\dfrac{\sec \left( x \right)\tan \left( x \right)}{1}} \right)=\underset{x\to \dfrac{\pi }{2}}{\mathop{\lim }}\,\left( \dfrac{{{\sec }^{2}}\left( x \right)}{\tan \left( x \right)}\times \dfrac{1}{\sec \left( x \right)\tan \left( x \right)} \right)$ after cancelling the common terms we have,
$\underset{x\to \dfrac{\pi }{2}}{\mathop{\lim }}\,\left( \dfrac{{{\sec }^{2}}\left( x \right)}{\tan \left( x \right)}\times \dfrac{1}{\sec \left( x \right)\tan \left( x \right)} \right)=\underset{x\to \dfrac{\pi }{2}}{\mathop{\lim }}\,\left( \dfrac{\sec \left( x \right)}{\tan \left( x \right)}\times \dfrac{1}{\tan \left( x \right)} \right)$
Now, we will convert this expression in terms of $\sin \left( x \right)$ and $\cos \left( x \right)$. As we know that the value of $\sec \left( x \right)=\dfrac{1}{\cos \left( x \right)}$ and the value of $\tan \left( x \right)=\dfrac{\sin \left( x \right)}{\cos \left( x \right)}$. Therefore, we have
$\underset{x\to \dfrac{\pi }{2}}{\mathop{\lim }}\,\left( \dfrac{\sec \left( x \right)}{\tan \left( x \right)}\times \dfrac{1}{\tan \left( x \right)} \right)=\underset{x\to \dfrac{\pi }{2}}{\mathop{\lim }}\,\left( \dfrac{\dfrac{1}{\cos \left( x \right)}}{\dfrac{\sin \left( x \right)}{\cos \left( x \right)}}\times \dfrac{1}{\dfrac{\sin \left( x \right)}{\cos \left( x \right)}} \right)$. Again by using the formula $\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}}=\dfrac{a}{b}\times \dfrac{d}{c}$ we have
$\underset{x\to \dfrac{\pi }{2}}{\mathop{\lim }}\,\left( \dfrac{\dfrac{1}{\cos \left( x \right)}}{\dfrac{\sin \left( x \right)}{\cos \left( x \right)}}\times \dfrac{1}{\dfrac{\sin \left( x \right)}{\cos \left( x \right)}} \right)=\underset{x\to \dfrac{\pi }{2}}{\mathop{\lim }}\,\left( \dfrac{1}{\cos \left( x \right)}\times \dfrac{\cos \left( x \right)}{\sin \left( x \right)}\times \dfrac{\cos \left( x \right)}{\sin \left( x \right)} \right)$
After cancelling the common terms we have $\underset{x\to \dfrac{\pi }{2}}{\mathop{\lim }}\,\left( \dfrac{1}{\cos \left( x \right)}\times \dfrac{\cos \left( x \right)}{\sin \left( x \right)}\times \dfrac{\cos \left( x \right)}{\sin \left( x \right)} \right)=\underset{x\to \dfrac{\pi }{2}}{\mathop{\lim }}\,\left( \dfrac{1}{\cos \left( x \right)}\times \dfrac{{{\cos }^{2}}\left( x \right)}{{{\sin }^{2}}\left( x \right)} \right)$
Thus, we have
$\underset{x\to \dfrac{\pi }{2}}{\mathop{\lim }}\,\left( \dfrac{1}{\cos \left( x \right)}\times \dfrac{{{\cos }^{2}}\left( x \right)}{{{\sin }^{2}}\left( x \right)} \right)=\underset{x\to \dfrac{\pi }{2}}{\mathop{\lim }}\,\left( \dfrac{\cos \left( x \right)}{{{\sin }^{2}}\left( x \right)} \right)$. Now, we will substitute the value of $\cos \left( \dfrac{\pi }{2} \right)=0$ and $\sin \left( \dfrac{\pi }{2} \right)=1$ therefore, we have $\underset{x\to \dfrac{\pi }{2}}{\mathop{\lim }}\,\left( \dfrac{\cos \left( x \right)}{{{\sin }^{2}}\left( x \right)} \right)=\left( \dfrac{\cos \left( \dfrac{\pi }{2} \right)}{{{\sin }^{2}}\left( \dfrac{\pi }{2} \right)} \right)$. After substituting the values we have $\left( \dfrac{\cos \left( \dfrac{\pi }{2} \right)}{{{\sin }^{2}}\left( \dfrac{\pi }{2} \right)} \right)=\dfrac{0}{1}$. This means that the value of the expression is equal to 0 after applying the limit.
Hence the value of $\underset{x\to \dfrac{\pi }{2}}{\mathop{\lim }}\,\left( \cos x\cdot \log \left( \tan x \right) \right)$is 0.

Note: While using L’Hopital’s rule one should notice that after first substitution of the limit we are getting the L’hopital’s signs or not. Here $0\cdot \infty $ is one of the signs by which we come to know that L’hopital’s rule is valid. One more thing should be noticed here and that is whenever we will differentiate any expression after L’hospital’s rule we should have a simpler term. That is, it should be in a fraction form and different functions should be on each other’s denominator and numerator. For example, we have converted $\cos x\cdot \log \left( \tan x \right)$ into $\dfrac{\log \left( \tan x \right)}{\sec \left( x \right)}$ and then differentiated.