
Evaluate the following limit: $\underset{x\to \dfrac{\pi }{2}}{\mathop \lim }\,\text{ }\dfrac{\tan 2x}{\left( x-\dfrac{\pi }{2} \right)}$
Answer
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Hint: In the above question of limit, first of all we will have to find the type of indeterminate form of limit for which we will put the limiting value of x in the given expression of the limit. By putting the limiting value of x , we get the form of the limit as (0/0). So, we will further use the L’hospital Rule to calculate the limits. L’Hospital Rule tells us that if we have an indeterminate form \[{}^{0}/{}_{0}\text{ or }{}^{\infty }/{}_{\infty }\] all we need to do is differentiate the numerator and differentiate the denominator and then take the limit.
Complete step-by-step answer:
In the above question if we put the limiting value of x in the expression to find the type of indeterminate form of the limit , we get :
\[\dfrac{\tan \left( 2\times \dfrac{\pi }{2} \right)}{\dfrac{\pi }{2}-\dfrac{\pi }{2}}=\dfrac{0}{0}\]
Here, we get the indeterminate form of the limit as (0/0).
So, we will use L’hospital rule to simplify the limit which is shown as belows:
\[\begin{align}
& \Rightarrow ~\underset{x\to \dfrac{\pi }{2}~}{\mathop \lim }\,\left( \dfrac{\dfrac{d}{dx}(\tan 2x)}{\dfrac{d}{dx}(x-\dfrac{\pi }{2})} \right) \\
& \Rightarrow \underset{x\to \dfrac{\pi }{2}~}{\mathop \lim }\,\left( \dfrac{{{\sec }^{2}}2x\times 2}{1} \right) \\
& \Rightarrow \underset{x\to \dfrac{\pi }{2}~}{\mathop \lim }\,\left( \dfrac{{{\sec }^{2}}\left( 2\times \dfrac{\pi }{2} \right)\times 2}{1} \right) \\
& \Rightarrow \underset{x\to \dfrac{\pi }{2}~}{\mathop \lim }\,\left( \dfrac{{{\sec }^{2}}\left( \pi \right)\times 2}{1} \right) \\
\end{align}\]
Since, the value of \[{{\sec }^{2}}\pi \] is equal to 1,
\[\Rightarrow \underset{x\to \dfrac{\pi }{2}}{\mathop \lim }\,\left( \dfrac{{{\sec }^{2}}\pi \times 2}{1} \right)=2\]
Therefore, the value of the limit is equal to 2.
Note: Just be careful while solving the limit, especially at the time of applying L’hospital Rule as there is a chance that you might make a mistake while solving it.
Also remember the L’Hospital Rule as it will help you a lot in these types of questions of limit.
Complete step-by-step answer:
In the above question if we put the limiting value of x in the expression to find the type of indeterminate form of the limit , we get :
\[\dfrac{\tan \left( 2\times \dfrac{\pi }{2} \right)}{\dfrac{\pi }{2}-\dfrac{\pi }{2}}=\dfrac{0}{0}\]
Here, we get the indeterminate form of the limit as (0/0).
So, we will use L’hospital rule to simplify the limit which is shown as belows:
\[\begin{align}
& \Rightarrow ~\underset{x\to \dfrac{\pi }{2}~}{\mathop \lim }\,\left( \dfrac{\dfrac{d}{dx}(\tan 2x)}{\dfrac{d}{dx}(x-\dfrac{\pi }{2})} \right) \\
& \Rightarrow \underset{x\to \dfrac{\pi }{2}~}{\mathop \lim }\,\left( \dfrac{{{\sec }^{2}}2x\times 2}{1} \right) \\
& \Rightarrow \underset{x\to \dfrac{\pi }{2}~}{\mathop \lim }\,\left( \dfrac{{{\sec }^{2}}\left( 2\times \dfrac{\pi }{2} \right)\times 2}{1} \right) \\
& \Rightarrow \underset{x\to \dfrac{\pi }{2}~}{\mathop \lim }\,\left( \dfrac{{{\sec }^{2}}\left( \pi \right)\times 2}{1} \right) \\
\end{align}\]
Since, the value of \[{{\sec }^{2}}\pi \] is equal to 1,
\[\Rightarrow \underset{x\to \dfrac{\pi }{2}}{\mathop \lim }\,\left( \dfrac{{{\sec }^{2}}\pi \times 2}{1} \right)=2\]
Therefore, the value of the limit is equal to 2.
Note: Just be careful while solving the limit, especially at the time of applying L’hospital Rule as there is a chance that you might make a mistake while solving it.
Also remember the L’Hospital Rule as it will help you a lot in these types of questions of limit.
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