Question

How do you find the limit of \[x\sin \left( \dfrac{\pi }{x} \right)\] as x approaches infinity?

Answer 1

Hint: Assume the given limit as ‘L’. Now, replace x with \[\dfrac{1}{y}\] and use the relation \[x\to \infty \] to deduce the relation \[y\to 0\]. Simplify the given limit and use the formula: - \[\underset{a\to 0}{\mathop{\lim }}\,\dfrac{\sin a}{a}=1\] to get the answer. The argument of sine function, i.e., the angle, and the denominator must be the same to use the formula.

Complete step by step answer:
Here, we have been provided with the expression \[x\sin \left( \dfrac{\pi }{x} \right)\] and we have been asked to find its limit as x is tending to infinity.
Now, let us assume the given limit as ‘L’. So, mathematically we can write the given expression as: -
\[\Rightarrow L=\underset{x\to \infty }{\mathop{\lim }}\,x\sin \left( \dfrac{\pi }{x} \right)\]
Replacing x with \[\dfrac{1}{y}\], we get,
\[\begin{align}
  & \because x\to \infty \\
 & \Rightarrow \dfrac{1}{y}\to \infty \\
\end{align}\]
Taking the reciprocal on both the sides, we get,
\[\begin{align}
  & \Rightarrow y\to \dfrac{1}{\infty } \\
 & \Rightarrow y\to 0 \\
\end{align}\]
Now, we can write the given expression of limit as: -
\[\Rightarrow L=\underset{y\to 0}{\mathop{\lim }}\,\dfrac{1}{y}\sin \left( \dfrac{\pi }{\dfrac{1}{y}} \right)\]
On simplifying we get,
\[\Rightarrow L=\underset{y\to 0}{\mathop{\lim }}\,\dfrac{\sin \left( \pi y \right)}{y}\]
We know that \[\underset{a\to 0}{\mathop{\lim }}\,\dfrac{\sin a}{a}=1\]. Here, we can see that the argument of the sine function and the denominator must be the same and both must be tending to zero to apply the formula. So, in \[\dfrac{\sin \left( \pi y \right)}{y}\] we need to change the denominator into ‘\[\pi y\]’. So, multiplying and dividing the expression of L with \[\pi \], we get,
\[\begin{align}
  & \Rightarrow L=\underset{y\to 0}{\mathop{\lim }}\,\dfrac{\sin \left( \pi y \right)}{y}\times \dfrac{\pi }{\pi } \\
 & \Rightarrow L=\underset{y\to 0}{\mathop{\lim }}\,\dfrac{\pi \sin \left( \pi y \right)}{\pi y} \\
\end{align}\]
As we can see that, \[y\to 0\], so multiplying both the sides with \[\pi \], we get,
\[\begin{align}
  & \Rightarrow \pi y\to 0\times \pi \\
 & \Rightarrow \pi y\to 0 \\
\end{align}\]
Therefore, we can write the given limit as,
\[\Rightarrow L=\underset{\pi y\to 0}{\mathop{\lim }}\,\dfrac{\pi \sin \left( \pi y \right)}{\left( \pi y \right)}\]
Since, ‘\[\pi \]’ is a constant so it can be taken out of the limit expression in the R.H.S. So, we get,
\[\Rightarrow L=\pi \left[ \underset{\pi y\to 0}{\mathop{\lim }}\,\dfrac{\sin \left( \pi y \right)}{\left( \pi y \right)} \right]\]
Applying the formula: - \[\underset{a\to 0}{\mathop{\lim }}\,\dfrac{\sin a}{a}=1\], we get,
\[\begin{align}
  & \Rightarrow L=\pi \times 1 \\
 & \Rightarrow L=\pi \\
\end{align}\]
Hence, the value of the given limit is \[\pi \].

Note:
 One may note that we cannot substitute the value of x equal to infinite to get the answer. This is because the given expression must be converted into \[\dfrac{0}{0}\] or \[\dfrac{\infty }{\infty }\] form first. Once this happens then only we can apply the required formula. You must remember some basic formulas of limits otherwise it will be difficult to solve the question. Note that once the given limit is converted into \[\dfrac{0}{0}\] or \[\dfrac{\infty }{\infty }\] form then we can apply L – Hospital’s Rule also to get the answer.