Question

Find the value of expression $\underset{x\to \infty }{\mathop{\lim }}\,\left( {{\tan }^{-1}}\dfrac{x+1}{x+4}-\dfrac{\pi }{4} \right)$ .

Answer 1

Hint: Write $\dfrac{\pi }{4}$ as ${{\tan }^{-1}}1$ and apply the trigonometric identity of \[{{\tan }^{-1}}A-{{\tan }^{-1}}B\] , which is given as \[{{\tan }^{-1}}A-{{\tan }^{-1}}B={{\tan }^{-1}}\left( \dfrac{A-B}{1+AB} \right)\]

Complete step-by-step answer:
Simplify the relation using above identity and now apply $x\to \infty $ to the updated expression and use result:
\[{{\tan }^{-1}}0=0\]

Let the limit of the given expression in the problem is ‘L’. So, we can write the equation as
$L=\underset{x\to \infty }{\mathop{\lim }}\,\left( {{\tan }^{-1}}\left( \dfrac{x+1}{x+4} \right)-\dfrac{\pi }{4} \right)................\left( i \right)$
Now, we can observe the expression $\left( \dfrac{x+1}{x+4} \right)$ will give form of the limit as $\dfrac{\infty }{\infty }$ if we put $x\to \infty $ to it. It means the given expression will be in indeterminate form for $x\to \infty $ . Hence, we need to simplify the expression further to get the limit of the given expression. So, as we know value of ${{\tan }^{-1}}\left( 1 \right)$ is $\dfrac{\pi }{4}$ . So, we can replace $\dfrac{\pi }{4}$ of the expression (i) by \[{{\tan }^{-1}}\left( 1 \right)\] . So, we can write expression (i) as
$L=\underset{x\to \infty }{\mathop{\lim }}\,\left( {{\tan }^{-1}}\left( \dfrac{x+1}{x+4} \right)-{{\tan }^{-1}}\left( 1 \right) \right)................\left( ii \right)$
Now, as we know the trigonometric identity of \[{{\tan }^{-1}}A-{{\tan }^{-1}}B\] can be given as
 \[{{\tan }^{-1}}A-{{\tan }^{-1}}B={{\tan }^{-1}}\left( \dfrac{A-B}{1+AB} \right)...........\left( iii \right)\]
Hence, we can apply the identity of equation (iii) to the equation (ii) and so, we can re-write the equation (ii) as
$\begin{align}
  & L=\underset{x\to \infty }{\mathop{\lim }}\,{{\tan }^{-1}}\left( \dfrac{\dfrac{x+1}{x+4}-1}{1+1\times \left( \dfrac{x+1}{x+4} \right)} \right) \\
 & \Rightarrow L=\underset{x\to \infty }{\mathop{\lim }}\,{{\tan }^{-1}}\left[ \dfrac{\left( \dfrac{\left( x+1 \right)-\left( x+4 \right)}{x+4} \right)}{\left( \dfrac{\left( x+4 \right)+\left( x+1 \right)}{\left( x+4 \right)} \right)} \right] \\
 & L=\underset{x\to \infty }{\mathop{\lim }}\,{{\tan }^{-1}}\left( \dfrac{x+1-x-4}{x+4+x+1} \right) \\
 & L=L=\underset{x\to \infty }{\mathop{\lim }}\,\left[ {{\tan }^{-1}}\left( \dfrac{-3}{2x+5} \right) \right] \\
\end{align}$
Now, we can put $x\to \infty $ to the function $\dfrac{-3}{2x+5}$ and hence, it will tend to 0, when $x\to \infty $ will be applied. Hence, we can get value of L as
$L={{\tan }^{-1}}\left( 0 \right)$
Now, we know the value of ${{\tan }^{-1}}\left( 0 \right)$ is given as 0. So, the value of ‘L’ will also be 0.
Hence, we get an answer as 0. So,
$\underset{x\to \infty }{\mathop{\lim }}\,\left[ {{\tan }^{-1}}\left( \dfrac{x+1}{x+4} \right)-\dfrac{\pi }{4} \right]=0$

Note: One may get confused with the term ${{\tan }^{-1}}\left( \dfrac{x+1}{x+4} \right)$ , which is giving an intermediate form and another term $\left( \dfrac{\pi }{4} \right)$ is a constant. So, one may think how a constant will change the form of an intermediate value. So, be clear with it that if any function is giving an intermediate form of limit then try to simplify the whole expression given to it, it will definitely change the characteristic of the limit value.
One may get confused with the identities ${{\tan }^{-1}}A+{{\tan }^{-1}}B$ and ${{\tan }^{-1}}A-{{\tan }^{-1}}B$ .
Both are given as
\[\begin{align}
  & {{\tan }^{-1}}A+{{\tan }^{-1}}B={{\tan }^{-1}}\left( \dfrac{A+B}{1-AB} \right) \\
 & {{\tan }^{-1}}A-{{\tan }^{-1}}B={{\tan }^{-1}}\left( \dfrac{A-B}{1+AB} \right) \\
\end{align}\]
Writing \[\dfrac{\pi }{4}\] to ${{\tan }^{-1}}1$ is the key point of the given problem.