Question

Find the limit as \[x\] approaches infinity of \[x\sin \left( {\dfrac{1}{x}} \right)\]

Answer 1

Hint: In order to solve this question, we first substitute \[\infty \] at the place of \[x\] then we see we end up with the indeterminate form of \[\infty \cdot 0\] .So, we need to rewrite this function so that it produces an indeterminate in the form of either \[\dfrac{\infty }{\infty }\] or \[\dfrac{0}{0}\] .After that we apply L-Hospital Rule which basically says to differentiate both numerator and denominator independently with respect to \[x\] .and then simplify it to get the result.

Complete step-by-step answer:
We are given a function, i.e., \[x\sin \left( {\dfrac{1}{x}} \right)\]
And we have to find the limit as \[x\] approaches infinity
i.e., we have to find \[\mathop {\lim }\limits_{x \to \infty } {\text{ }}x\sin \left( {\dfrac{1}{x}} \right)\]
So, first we substitute \[\infty \] at the place of \[x\] we get
\[ \Rightarrow \mathop {\lim }\limits_{x \to \infty } {\text{ x}}\sin \left( {\dfrac{1}{x}} \right) = \infty \sin \left( {\dfrac{1}{\infty }} \right)\]
We know that \[\dfrac{1}{\infty } = 0\]
\[\therefore \mathop {\lim }\limits_{x \to \infty } {\text{ }}x\sin \left( {\dfrac{1}{x}} \right) = \infty \sin \left( 0 \right) = \infty \cdot 0\]
Here, we see we end up with the indeterminate form of \[\infty \cdot 0\] . So, we need to rewrite this function so that it produces an indeterminate in the form of either \[\dfrac{\infty }{\infty }\] or \[\dfrac{0}{0}\]
So, we can write \[x\sin \left( {\dfrac{1}{x}} \right)\] as \[\dfrac{{\sin \left( {\dfrac{1}{x}} \right)}}{{{x^{ - 1}}}}\]
\[ \Rightarrow \mathop {\lim }\limits_{x \to \infty } {\text{ }}\dfrac{{\sin \left( {\dfrac{1}{x}} \right)}}{{{x^{ - 1}}}} = \mathop {\lim }\limits_{x \to \infty } \dfrac{{\sin \left( {\dfrac{1}{x}} \right)}}{{\dfrac{1}{x}}}\]
On substituting \[\infty \] at the place of \[x\] we get,
\[ \Rightarrow \mathop {\lim }\limits_{x \to \infty } {\text{ }}\dfrac{{\sin \left( {\dfrac{1}{x}} \right)}}{{\dfrac{1}{x}}} = \dfrac{{\sin \left( {\dfrac{1}{\infty }} \right)}}{{\dfrac{1}{\infty }}}\]
Now we know that \[\dfrac{1}{\infty } = 0\]
\[ \Rightarrow \mathop {\lim }\limits_{x \to \infty } {\text{ }}\dfrac{{\sin \left( {\dfrac{1}{x}} \right)}}{{\dfrac{1}{x}}}{\text{ = }}\dfrac{{\sin \left( 0 \right)}}{0} = \dfrac{0}{0}\] which is an indeterminate form of \[\dfrac{0}{0}\]
So, now we apply L-Hospital Rule:
which basically says to differentiate both numerator and denominator independently with respect to \[x\]
Therefore, on applying L-Hospital Rule,
\[\mathop {\lim }\limits_{x \to \infty } {\text{ }}\dfrac{{\dfrac{d}{{dx}}\left( {\sin \left( {\dfrac{1}{x}} \right)} \right)}}{{\dfrac{d}{{dx}}\left( {\dfrac{1}{x}} \right)}}\]
We know that,
\[\dfrac{{d\left( {\sin x} \right)}}{{dx}} = \cos x\] and \[\dfrac{{d\left( {\dfrac{1}{x}} \right)}}{{dx}} = \dfrac{{ - 1}}{{{x^2}}} = \left( { - 1} \right){x^{ - 2}}\]
Therefore, we get
\[\mathop {\lim }\limits_{x \to \infty } {\text{ }}\dfrac{{\cos \left( {\dfrac{1}{x}} \right)\left( { - 1} \right){x^{ - 2}}}}{{\left( { - 1} \right){x^{ - 2}}}}\]
On cancelling \[\left( { - 1} \right){x^{ - 2}}\] from numerator and denominator, we get
\[\mathop {\lim }\limits_{x \to \infty } {\text{ }}\cos \left( {\dfrac{1}{x}} \right)\]
On substituting \[\infty \] at the place of \[x\] we get,
\[\mathop {\lim }\limits_{x \to \infty } {\text{ }}\cos \left( {\dfrac{1}{x}} \right) = \cos \left( {\dfrac{1}{\infty }} \right)\]
Now we know that \[\dfrac{1}{\infty } = 0\]
\[\mathop {\lim }\limits_{x \to \infty } {\text{ }}\cos \left( {\dfrac{1}{x}} \right) = \cos \left( 0 \right)\]
We know that \[\cos \left( 0 \right) = 1\]
So, \[\mathop {\lim }\limits_{x \to \infty } {\text{ }}\cos \left( {\dfrac{1}{x}} \right) = 1\] which is our required answer.
Hence, \[\mathop {\lim }\limits_{x \to \infty } {\text{ }}x\sin \left( {\dfrac{1}{x}} \right) = 1\]
So, the correct answer is “1”.

Note: Instead of L-Hospital Rule, one can use the fundamental trigonometric limit: \[\mathop {\lim }\limits_{h \to 0} {\text{ }}\dfrac{{\sinh }}{h} = 1\]
So, the limit given can be written as:
\[\mathop {\lim }\limits_{x \to \infty } {\text{ }}\dfrac{{\sin \left( {\dfrac{1}{x}} \right)}}{{\left( {\dfrac{1}{x}} \right)}}\]
Now, as \[x \to \infty \] , we know that \[\dfrac{1}{x} \to 0\]
Therefore, we get
\[\mathop {\lim }\limits_{\dfrac{1}{x} \to 0} {\text{ }}\dfrac{{\sin \left( {\dfrac{1}{x}} \right)}}{{\left( {\dfrac{1}{x}} \right)}}\]
Now, with \[h = \dfrac{1}{x}\]
This becomes, \[\mathop {\lim }\limits_{h \to 0} {\text{ }}\dfrac{{\sinh }}{h}\] which is \[1\]