What is the value of $ \underset{x\to \infty }{\mathop{\lim }}\,{{x}^{2}}\sin \left( \dfrac{1}{x} \right) $ ?

Hint: We first change the variable of the limit with the conversion of $ \dfrac{1}{x}=z $ . The limit value of the function also changes from $ x\to \infty $ to $ z\to 0 $ . We change all the variables from $ x $ to $ z $ as $ \underset{x\to \infty }{\mathop{\lim }}\,{{x}^{2}}\sin \left( \dfrac{1}{x} \right)=\underset{z\to 0}{\mathop{\lim }}\,\dfrac{\sin z}{{{z}^{2}}} $ . We use the theorems $ \underset{x\to a}{\mathop{\lim }}\,f\left( x \right)g\left( x \right)=\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)\times \underset{x\to a}{\mathop{\lim }}\,g\left( x \right) $ and $ \underset{z\to 0}{\mathop{\lim }}\,\dfrac{\sin z}{z}=1 $ . We put the values to get $ \underset{x\to \infty }{\mathop{\lim }}\,{{x}^{2}}\sin \left( \dfrac{1}{x} \right)=\infty $ .

Complete step by step solution:
First, we try to change the variable for the given limit value of $ \underset{x\to \infty }{\mathop{\lim }}\,{{x}^{2}}\sin \left( \dfrac{1}{x} \right) $ .
We use the conversion of $ \dfrac{1}{x}=z $ . The given limit form is $ x\to \infty $ .
Due to the change of the variable the limit also changes to $ z=\dfrac{1}{x}\to 0 $ .
The value of the $ x $ becomes $ x=\dfrac{1}{z} $ .
The function changes to $ \underset{x\to \infty }{\mathop{\lim }}\,{{x}^{2}}\sin \left( \dfrac{1}{x} \right)=\underset{z\to 0}{\mathop{\lim }}\,\dfrac{1}{{{z}^{2}}}\sin z=\underset{z\to 0}{\mathop{\lim }}\,\dfrac{\sin z}{{{z}^{2}}} $ .
Now we are going to apply the limit theorem of $ \underset{x\to a}{\mathop{\lim }}\,f\left( x \right)g\left( x \right)=\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)\times \underset{x\to a}{\mathop{\lim }}\,g\left( x \right) $ .
Therefore, $ \underset{z\to 0}{\mathop{\lim }}\,\dfrac{\sin z}{{{z}^{2}}}=\underset{z\to 0}{\mathop{\lim }}\,\left( \dfrac{\sin z}{z}\times \dfrac{1}{z} \right)=\underset{z\to 0}{\mathop{\lim }}\,\dfrac{\sin z}{z}\times \underset{z\to 0}{\mathop{\lim }}\,\dfrac{1}{z} $ .
Now we have the limit theorem $ \underset{z\to 0}{\mathop{\lim }}\,\dfrac{\sin z}{z}=1 $ .
We apply the theorem to get $ \underset{z\to 0}{\mathop{\lim }}\,\dfrac{\sin z}{z}\times \underset{z\to 0}{\mathop{\lim }}\,\dfrac{1}{z}=\underset{z\to 0}{\mathop{\lim }}\,\dfrac{1}{z} $ .
Now we apply the limit value to get $ \underset{z\to 0}{\mathop{\lim }}\,\dfrac{1}{z}=\infty $ .
Therefore, $ \underset{x\to \infty }{\mathop{\lim }}\,{{x}^{2}}\sin \left( \dfrac{1}{x} \right)=\underset{z\to 0}{\mathop{\lim }}\,\dfrac{\sin z}{{{z}^{2}}}=\infty $ .
The limit value of $ \underset{x\to \infty }{\mathop{\lim }}\,{{x}^{2}}\sin \left( \dfrac{1}{x} \right) $ is $ \infty $ .
So, the correct answer is “ $ \infty $”.

Note: The precise definition of a limit is something we use as a proof for the existence of a limit. When we’re evaluating a limit, we’re looking at the function as it approaches a specific point. we approach a particular value of x, the function itself gets closer and closer to a particular value.