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 $1$